Q1 MHT CET 2026 · 13 MAY · SHIFT 1 Easy ☆
Junction Transistor · Structure & biasing
An n-p-n transistor can be considered to be equivalent to two diodes connected. Which one of the following figure is correct?
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Answer: (b)
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Q2 MHT CET 2026 · 19 APR · SHIFT 2 Easy ☆
Junction Transistor · Structure & biasing
In a transistor, comparing the doping of emitter, base and collector, the part which is heavily doped and that which is lightly doped are respectively
(a) collector and emitter (b) emitter and base (c) collector and base (d) emitter and collector Show answer and solution
Answer: (b)
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Q3 MHT CET 2026 · 14 MAY · SHIFT 1 Easy ☆
Junction Transistor · Current relations
For a transistor, the current ratio α D C = 69 70 \alpha_{DC} = \frac{69}{70} α D C = 70 69 . The current gain β D C \beta_{DC} β D C is
(a) 66 66 66 (b) 67 67 67 (c) 69 69 69 (d) 71 71 71 Show answer and solution
Answer: (c)
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Q4 MHT CET 2026 · 13 APR · SHIFT 2 Easy ☆
Junction Transistor · Current relations
If transistor having α = 0.9 \alpha = 0.9 α = 0.9 , is used in CE configuration, then for a change of 0.4 0.4 0.4 mA in base current, what will be the change in collector current?
(a) 3.6 3.6 3.6 mA(b) 4 4 4 mA(c) 0.9 0.9 0.9 mA(d) 36 36 36 mAShow answer and solution
Answer: (a)
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Q5 MHT CET 2026 · 13 APR · SHIFT 1 Easy ☆
Junction Transistor · Current relations
In transistor amplifier, base emitter junction is forward biased and collector base junction is reversed biased. The current gain is
(a) Δ I C Δ I B \frac{\Delta I_C}{\Delta I_B} Δ I B Δ I C (b) Δ I B Δ I C \frac{\Delta I_B}{\Delta I_C} Δ I C Δ I B (c) Δ I E Δ I B \frac{\Delta I_E}{\Delta I_B} Δ I B Δ I E (d) Δ I B Δ I E \frac{\Delta I_B}{\Delta I_E} Δ I E Δ I B Show answer and solution
Answer: (a)
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Q6 MHT CET 2026 · 14 MAY · SHIFT 2 Easy ☆
Junction Transistor · CE amplifier gains
The current gain in the common emitter mode of a transistor is 10 10 10 . The input impedance is 20 k Ω 20\,k\Omega 20 k Ω and load resistance 100 k Ω 100\,k\Omega 100 k Ω . The power gain is
(a) 300 300 300 (b) 500 500 500 (c) 200 200 200 (d) 100 100 100 Show answer and solution
Answer: (b)
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Q7 MHT CET 2026 · 14 MAY · SHIFT 1 Easy ☆
Junction Transistor · CE amplifier gains
β , R L \beta, R_L β , R L and r r r are the a.c. current gain, load resistance and the input resistance of a transistor respectively in common emitter configuration, the voltage gain and the power gain are respectively
(a) β ( R L / r ) \beta(R_L/r) β ( R L / r ) and β 2 ( R L / r ) \beta^2(R_L/r) β 2 ( R L / r ) (b) β ( r / R L ) \beta(r/R_L) β ( r / R L ) and β 2 ( r / R L ) \beta^2(r/R_L) β 2 ( r / R L ) (c) β ( R L / r ) \beta(R_L/r) β ( R L / r ) and β ( R L / r ) 2 \beta(R_L/r)^2 β ( R L / r ) 2 (d) β ( r / R L ) \beta(r/R_L) β ( r / R L ) and β ( r / R L ) 2 \beta(r/R_L)^2 β ( r / R L ) 2 Show answer and solution
Answer: (a)
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Q8 MHT CET 2026 · 13 MAY · SHIFT 1 Easy ☆
Junction Transistor · CE amplifier gains
In common emitter amplifier, using output resistance 5000 Ω 5000\,\Omega 5000 Ω and input resistance 2000 Ω 2000\,\Omega 2000 Ω , if the peak value of input signal voltage is 10 10 10 mV and β = 50 \beta = 50 β = 50 , then peak value of output voltage is
(a) 5 × 1 0 − 6 5 \times 10^{-6} 5 × 1 0 − 6 V(b) 12.50 × 1 0 − 6 12.50 \times 10^{-6} 12.50 × 1 0 − 6 V(c) 1.25 1.25 1.25 V(d) 125.0 125.0 125.0 VShow answer and solution
Answer: (c)
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Q9 MHT CET 2026 · 20 APR · SHIFT 1 Easy ☆
Junction Transistor · CE amplifier gains
In a common emitter transistor amplifier, AC current gain is 65 65 65 , the load resistance is 5400 Ω 5400\,\Omega 5400 Ω and input resistance of the transistor is 450 Ω 450\,\Omega 450 Ω . The voltage gain is
(a) 460 460 460 (b) 540 540 540 (c) 780 780 780 (d) 7800 7800 7800 Show answer and solution
Answer: (c)
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Q10 MHT CET 2026 · 17 APR · SHIFT 2 Easy ☆
Junction Transistor · CE amplifier gains
In common emitter configuration of transistor amplifier, r i , R L r_i, R_L r i , R L and β \beta β represent the input resistance, load resistance and the a.c. current gain respectively. The voltage gain A V A_V A V and power gain A P A_P A P are represented in magnitude respectively by
(a) β ( r i R L ) , β 2 ( r i R L ) \beta\left(\frac{r_i}{R_L}\right), \beta^2\left(\frac{r_i}{R_L}\right) β ( R L r i ) , β 2 ( R L r i ) (b) β ( R L r i ) , β ( R L r i ) 2 \beta\left(\frac{R_L}{r_i}\right), \beta\left(\frac{R_L}{r_i}\right)^2 β ( r i R L ) , β ( r i R L ) 2 (c) β ( R L r i ) , β 2 ( R L r i ) \beta\left(\frac{R_L}{r_i}\right), \beta^2\left(\frac{R_L}{r_i}\right) β ( r i R L ) , β 2 ( r i R L ) (d) β ( r i R L ) , β ( r i R L ) 2 \beta\left(\frac{r_i}{R_L}\right), \beta\left(\frac{r_i}{R_L}\right)^{2} β ( R L r i ) , β ( R L r i ) 2 Show answer and solution
Answer: (c)
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Q11 MHT CET 2026 · 13 MAY · SHIFT 2 Medium ☆
Junction Transistor · Current relations
In an n-p-n transistor the collector current is 9 9 9 mA. If 90 % 90\% 90% of the electrons from the emitter reach the collector then
(a) β = 9.0 \beta = 9.0 β = 9.0 (b) the base current is 10 10 10 mA (c) the emitter current is 1 1 1 mA (d) β = 0.9 \beta = 0.9 β = 0.9 Show answer and solution
Answer: (a)
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Q12 MHT CET 2026 · 18 APR · SHIFT 2 Medium ☆
Junction Transistor · Current relations
In an NPN transistor, the collector current is 28 mA. If 80% electrons reach the collector, its base current in mA is
(a) 7 7 7 mA(b) 14 14 14 mA(c) 28 28 28 mA(d) 35 35 35 mAShow answer and solution
Answer: (a)
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Q13 MHT CET 2026 · 13 APR · SHIFT 1 Medium ☆
Junction Transistor · Current relations
In common emitter mode of transistor, the d.c. current gain is 20, the emitter current is 7 mA. The collector current is
(a) 20 3 \frac{20}{3} 3 20 mA(b) 14 5 \frac{14}{5} 5 14 mA(c) 1 3 \frac{1}{3} 3 1 mA(d) 8 13 \frac{8}{13} 13 8 mAShow answer and solution
Answer: (a)
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Q14 MHT CET 2026 · 11 APR · SHIFT 2 Medium ☆
Junction Transistor · Current relations
In common emitter mode of transistor, the d.c. current gain is 20 20 20 , the emitter current is 7 7 7 mA. The collector current is
(a) 14 3 \frac{14}{3} 3 14 mA(b) 20 3 \frac{20}{3} 3 20 mA(c) 7 3 \frac{7}{3} 3 7 mA(d) 8 3 \frac{8}{3} 3 8 mAShow answer and solution
Answer: (b)
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Q15 MHT CET 2026 · 16 APR · SHIFT 1 Medium ☆
Junction Transistor · CE amplifier gains
In a transistor amplifier, a change of 0.2 mA in the base current causes a change of 5 mA in the collector current. If input resistance is 2 2 2 kΩ \Omega Ω and voltage gain is 75, the load resistance used in the circuit is
(a) 4 4 4 kΩ \Omega Ω (b) 6 6 6 kΩ \Omega Ω (c) 8 8 8 kΩ \Omega Ω (d) 2 2 2 kΩ \Omega Ω Show answer and solution
Answer: (b)
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Q16 MHT CET 2026 · 15 APR · SHIFT 2 Medium ☆
Junction Transistor · CE amplifier gains
In common emitter amplifier, a change of 0.2 0.2 0.2 mA in the base current causes a change of 5 5 5 mA in the collector current. If input resistance is 2 2 2 kΩ \Omega Ω and voltage gain is 75 75 75 , the load resistance used in the circuit is
(a) 2 2 2 kΩ \Omega Ω (b) 3 3 3 kΩ \Omega Ω (c) 4 4 4 kΩ \Omega Ω (d) 6 6 6 kΩ \Omega Ω Show answer and solution
Answer: (d)
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Q17 MHT CET 2026 · 11 APR · SHIFT 1 Medium ☆
Junction Transistor · CE amplifier gains
In common emitter transistor amplifier, the output resistance is 500 500 500 kΩ \Omega Ω and the current gain β = 50 \beta = 50 β = 50 . If power gain of amplifier is 5 × 1 0 6 5 \times 10^{6} 5 × 1 0 6 , what is the input resistance?
(a) 325 Ω 325\ \Omega 325 Ω (b) 150 Ω 150\ \Omega 150 Ω (c) 350 Ω 350\ \Omega 350 Ω (d) 250 Ω 250\ \Omega 250 Ω Show answer and solution
Answer: (d)
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Q18 MHT CET 2026 · 15 MAY · SHIFT 1 Medium ☆
Junction Transistor · Characteristics & transistor testing
In an n-p-n transistor in common emitter configuration when V B E V_{BE} V BE is increased by small amount then
(a) I B I_B I B increases and I C I_C I C increases proportionally(b) I B I_B I B increases and I C I_C I C remain constant(c) I B I_B I B remain constant and I C I_C I C increases(d) both I B I_B I B and I C I_C I C remain nearly constant Show answer and solution
Answer: (a)
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Q19 MHT CET 2025 · 26 APR · SHIFT 2 Easy ☆
Junction Transistor · Current relations
For a transistor, α dc \alpha_\text{dc} α dc and β dc \beta_\text{dc} β dc are the current ratios, then the value of β dc − α dc α dc × β dc \frac{\beta_\text{dc} - \alpha_\text{dc}}{\alpha_\text{dc} \times \beta_\text{dc}} α dc × β dc β dc − α dc
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Answer: (d)
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Q20 MHT CET 2025 · 23 APR · SHIFT 1 Easy ☆
Junction Transistor · Current relations
In a common emitter transistor amplifier circuit, the current gain α = 0.8 \alpha = 0.8 α = 0.8 . If the base current changes by 3 μ A 3\,\mu\text{A} 3 μ A then the collector current changes by
(a) 4 μ A 4\,\mu\text{A} 4 μ A (b) 12 μ A 12\,\mu\text{A} 12 μ A (c) 6 μ A 6\,\mu\text{A} 6 μ A (d) 24 μ A 24\,\mu\text{A} 24 μ A Show answer and solution
Answer: (b)
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Q21 MHT CET 2025 · 23 APR · SHIFT 1 Easy ☆
Junction Transistor · Current relations
In the case of common emitter transistor amplifier, the ratio of collector current to emitter current is 0.96, then the current gain ( β ) (\beta) ( β ) of the amplifier is
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Answer: (c)
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Q22 MHT CET 2025 · 19 APR · SHIFT 1 Easy ☆
Junction Transistor · Current relations
For a common emitter transistor, if I C I E = 0.95 \frac{I_\text{C}}{I_\text{E}} = 0.95 I E I C = 0.95 , then the current gain is
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Answer: (c)
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Q23 MHT CET 2025 · 19 APR · SHIFT 1 Easy ☆
Junction Transistor · Current relations
In the case of constant ' α \alpha α ' and ' β \beta β ' of a transistor ( α \alpha α and β \beta β are current ratios)
(a) β < 1 , α > 1 \beta < 1, \alpha > 1 β < 1 , α > 1 (b) β > 1 , α < 1 \beta > 1, \alpha < 1 β > 1 , α < 1 (c) α = β \alpha = \beta α = β (d) α = β 2 \alpha = \beta^2 α = β 2 Show answer and solution
Answer: (b)
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Q24 MHT CET 2025 · 22 APR · SHIFT 1 Easy ☆
Junction Transistor · CE amplifier gains
In a common emitter transistor amplifier, the output voltage and input voltage have a phase difference of
(a) 0 c 0^{c} 0 c (b) π c 2 \frac{\pi^{c}}{2} 2 π c (c) 3 π c 4 \frac{3\pi^{c}}{4} 4 3 π c (d) π c \pi^{c} π c Show answer and solution
Answer: (d)
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Q25 MHT CET 2025 · 21 APR · SHIFT 2 Easy ☆
Junction Transistor · CE amplifier gains
In common emitter mode of a transistor, the current gain is 8. The input impedance is 25 k Ω 25\text{ k}\Omega 25 k Ω and load resistance is 75 k Ω 75\text{ k}\Omega 75 k Ω . The power gain is
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Answer: (a)
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Q26 MHT CET 2025 · 20 APR · SHIFT 1 Easy ☆
Junction Transistor · CE amplifier gains
When an n-p-n junction transistor is used as an amplifier in common emitter mode,
(a) the base emitter junction is reverse biased. (b) the input impedance is high and the output impedance is low. (c) input signal is applied between the base and emitter and the output is obtained between collector and emitter. (d) the input voltage and the output voltage are in phase. Show answer and solution
Answer: (c)
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Q27 MHT CET 2025 · 19 APR · SHIFT 2 Easy ☆
Junction Transistor · CE amplifier gains
In a transistor amplifier, AC current gain is 64, the load resistance is 5400 Ω 5400\ \Omega 5400 Ω and the input resistance of the transistor is 540 Ω 540\ \Omega 540 Ω . The voltage gain is
(a) 540 (b) 600 (c) 640 (d) 6400 Show answer and solution
Answer: (c)
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Q28 MHT CET 2025 · 21 APR · SHIFT 1 Medium ☆
Junction Transistor · Current relations
In an n-p-n transistor amplifier, the collector current is 8 mA. If 80 % 80\% 80% of the electrons from the emitter reach the collector, then ( α \alpha α and β \beta β are current ratios)
(a) the base current is 20 mA. (b) the emitter current is 2 mA. (c) α = 0.8 , β = 4.0 \alpha = 0.8, \beta = 4.0 α = 0.8 , β = 4.0 (d) α = 4.0 , β = 0.8 \alpha = 4.0, \beta = 0.8 α = 4.0 , β = 0.8 Show answer and solution
Answer: (c)
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Q29 MHT CET 2025 · 20 APR · SHIFT 2 Medium ☆
Junction Transistor · Current relations
In an n-p-n transistor amplifier, the collector current is 9 mA. If 90 % 90\% 90% of the electrons emitted reach the collector, then
(a) α = 9 , β = 0.9 \alpha = 9, \beta = 0.9 α = 9 , β = 0.9 (b) α = 0.9 , β = 90 \alpha = 0.9, \beta = 90 α = 0.9 , β = 90 (c) α = 0.9 , β = 9 \alpha = 0.9, \beta = 9 α = 0.9 , β = 9 (d) α = 90 , β = 0.9 \alpha = 90, \beta = 0.9 α = 90 , β = 0.9 Show answer and solution
Answer: (c)
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Q30 MHT CET 2025 · 26 APR · SHIFT 1 Medium ☆
Junction Transistor · CE amplifier gains
In a common emitter transistor amplifier circuit, the input resistance is 1.8 k Ω 1.8\text{ k}\Omega 1.8 k Ω and output is obtained across a load resistance of 9 k Ω 9\text{ k}\Omega 9 k Ω . The alternating current gain is 70. Corresponding to an a.c. input voltage of 6 mV, the output voltage will be
(a) 0.7 V (b) 1.4 V (c) 2.1 V (d) 4.2 V Show answer and solution
Answer: (c)
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Q31 MHT CET 2025 · 25 APR · SHIFT 2 Medium ☆
Junction Transistor · CE amplifier gains
In a common emitter amplifier configuration, the current gain is 62. The collector resistance and input resistance are 5 k Ω 5\text{ k}\Omega 5 k Ω and 500 Ω 500\ \Omega 500 Ω respectively. If the input voltage is 0.01 V, the output voltage will be
(a) 0.62 V (b) 6.2 V (c) 62 V (d) 620 V Show answer and solution
Answer: (b)
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Q32 MHT CET 2025 · 20 APR · SHIFT 1 Medium ☆
Junction Transistor · CE amplifier gains
In a transistor (common emitter configuration) the ratio of power gain to voltage gain is ( α \alpha α and β \beta β are current ratios)
(a) β α \frac{\beta}{\alpha} α β (b) α β \alpha\beta α β (c) α \alpha α (d) β \beta β Show answer and solution
Answer: (d)
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Q33 MHT CET 2024 · 3 MAY · SHIFT 1 Easy ☆
Junction Transistor · Structure & biasing
An n-p-n transistor can be considered to be equivalent to two diodes connected. The correct figure out of the following is
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Answer: (b)
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Q34 MHT CET 2024 · 15 MAY · SHIFT 1 Easy ☆
Junction Transistor · Current relations
For a transistor, current gain ( β ) = 50 (\beta) = 50 ( β ) = 50 . To change the collector current by 350 μ A 350\,\mu\text{A} 350 μ A , the base current should be changed by
(a) ( 50 350 ) μ A \left(\frac{50}{350}\right)\mu\text{A} ( 350 50 ) μ A (b) ( 350 − 50 ) μ A (350 - 50)\,\mu\text{A} ( 350 − 50 ) μ A (c) ( 350 + 50 ) μ A (350 + 50)\,\mu\text{A} ( 350 + 50 ) μ A (d) ( 350 50 ) μ A \left(\frac{350}{50}\right)\mu\text{A} ( 50 350 ) μ A Show answer and solution
Answer: (d)
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Q35 MHT CET 2024 · 16 MAY · SHIFT 1 Easy ☆
Junction Transistor · CE amplifier gains
The input signal given to C.E. amplifier having a voltage gain of 126 is V i = 2 cos ( 12 t + π 3 ) V_i = 2\cos\left(12t + \frac{\pi}{3}\right) V i = 2 cos ( 12 t + 3 π ) . The corresponding output signal will be
(a) 252 cos ( 12 t + 4 π 3 ) 252\cos\left(12t + \frac{4\pi}{3}\right) 252 cos ( 12 t + 3 4 π ) (b) 252 cos ( 12 t + π 3 ) 252\cos\left(12t + \frac{\pi}{3}\right) 252 cos ( 12 t + 3 π ) (c) 63 cos ( 12 t + 2 π 3 ) 63\cos\left(12t + \frac{2\pi}{3}\right) 63 cos ( 12 t + 3 2 π ) (d) 2 cos ( 12 t + 5 π 3 ) 2\cos\left(12t + \frac{5\pi}{3}\right) 2 cos ( 12 t + 3 5 π ) Show answer and solution
Answer: (a)
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Q36 MHT CET 2024 · 11 MAY · SHIFT 2 Easy ☆
Junction Transistor · CE amplifier gains
When n-p-n junction transistor is used as an amplifier in common emitter mode,
(a) the input impedance is high and the output impedance is low. (b) the base emitter junction is reverse biased. (c) input signal is applied between the base and emitter and output is obtained between collector and emitter. (d) the input voltage and output voltage are in phase. Show answer and solution
Answer: (c)
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Q37 MHT CET 2024 · 3 MAY · SHIFT 1 Easy ☆
Junction Transistor · CE amplifier gains
In common emitter transistor amplifier, load resistance is 6.5 k Ω 6.5\text{ k}\Omega 6.5 k Ω and an input resistance is 1.3 k Ω 1.3\text{ k}\Omega 1.3 k Ω . If the current gain is 78, the voltage gain is
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Answer: (b)
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Q38 MHT CET 2024 · 4 MAY · SHIFT 1 Medium ☆
Junction Transistor · Current relations
The collector supply voltage is 6 V and a voltage drop across a resistor of 600 Ω 600\ \Omega 600 Ω in the collector circuit is 0.6 V, in a circuit of transistor connected in common emitter mode. If the current gain is 20 then the base current is
(a) 0.25 mA (b) 0.05 mA (c) 0.12 mA (d) 0.02 mA Show answer and solution
Answer: (b)
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Q39 MHT CET 2024 · 2 MAY · SHIFT 1 Medium ☆
Junction Transistor · Current relations
In an NPN transistor 1 0 10 10^{10} 1 0 10 electrons enter the emitter in 1 0 − 6 s 10^{-6}\text{ s} 1 0 − 6 s and 2 % 2\% 2% electrons recombine with holes in base. The current ratios ' α \alpha α ' and ' β \beta β ' of a transistor are respectively (nearly)
(a) 0.98 , 49 0.98, 49 0.98 , 49 (b) 49 , 0.98 49, 0.98 49 , 0.98 (c) 0.49 , 98 0.49, 98 0.49 , 98 (d) 98 , 0.49 98, 0.49 98 , 0.49 Show answer and solution
Answer: (a)
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Q40 MHT CET 2024 · 9 MAY · SHIFT 1 Medium ☆
Junction Transistor · CE amplifier gains
The current amplification factor of a transistor is 50. The input resistance when used in common emitter mode is 1 k Ω 1\text{ k}\Omega 1 k Ω . The peak value for an a.c. input voltage of 0.01 V peak is
(a) 100 μ A 100\ \mu\text{A} 100 μ A (b) 0.01 mA (c) 0.25 mA (d) 500 μ A 500\ \mu\text{A} 500 μ A Show answer and solution
Answer: (d)
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Q41 JEE Main 2023 · 31 JAN · SHIFT 2 Easy Old syllabus ☆
Junction Transistor · Structure & biasing
Given below are two statements:
Statement I: In a typical transistor, all three regions emitter, base and collector have same doping level.
Statement II: In a transistor, collector is the thickest and base is the thinnest segment. In the light of the above statements, choose the most appropriate answer from the options given below.
(a) Statement I is correct but Statement II is incorrect (b) Both Statement I and Statement II are incorrect (c) Statement I is incorrect but Statement II is correct (d) Both Statement I and Statement II are correct Show answer and solution
Answer: (c)
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Q42 MHT CET 2023 · 14 MAY · SHIFT 1 Easy ☆
Junction Transistor · Current relations
For a common emitter configuration, if 'α \alpha α ' and 'β \beta β ' have their usual meanings, the incorrect relation between 'α \alpha α ' and 'β \beta β ' is
(a) 1 α = 1 β + 1 \frac{1}{\alpha} = \frac{1}{\beta} + 1 α 1 = β 1 + 1 (b) α = β 1 − β \alpha = \frac{\beta}{1-\beta} α = 1 − β β (c) α = β 1 + β \alpha = \frac{\beta}{1+\beta} α = 1 + β β (d) 1 β = 1 α − 1 \frac{1}{\beta} = \frac{1}{\alpha} - 1 β 1 = α 1 − 1 Show answer and solution
Answer: (b)
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Q43 MHT CET 2023 · 11 MAY · SHIFT 2 Easy ☆
Junction Transistor · Current relations
In the study of transistor as an amplifier if α = I C I E = 0.98 \alpha = \frac{I_C}{I_E} = 0.98 α = I E I C = 0.98 and β = I C I B = 49 \beta = \frac{I_C}{I_B} = 49 β = I B I C = 49 , where I C , I B I_C, I_B I C , I B and I E \text{I}_\text{E} I E are collector, base and emitter current respectively then ( 1 α − 1 β ) \left(\frac{1}{\alpha} - \frac{1}{\beta}\right) ( α 1 − β 1 ) is equal to
(a) zero (b) 1 2 \frac{1}{2} 2 1 (c) 2 (d) 1 Show answer and solution
Answer: (d)
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Q44 JEE Main 2023 · 12 APR · SHIFT 1 Easy Old syllabus ☆
Junction Transistor · Current relations
In an n-p-n common emitter (CE) transistor the collector current changes from 5 mA \text{mA} mA to 16 mA 16\text{ mA} 16 mA for the change in base current from 100 μ A 100\,\mu\text{A} 100 μ A and 200 μ A 200\,\mu\text{A} 200 μ A , respectively. The current gain of transistor is __________.
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Answer: (d)
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Q45 MHT CET 2023 · 9 MAY · SHIFT 2 Easy ☆
Junction Transistor · CE amplifier gains
In a transistor, in common emitter configuration, the ratio of power gain to voltage gain is
(a) α \alpha α (b) β α \frac{\beta}{\alpha} α β (c) β α \beta\alpha β α (d) β \beta β Show answer and solution
Answer: (d)
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Q46 MHT CET 2023 · 10 MAY · SHIFT 2 Medium ☆
Junction Transistor · Structure & biasing
Which one of the operations of n − p − n \mathrm{n}-\mathrm{p}-\mathrm{n} n − p − n transistor differs from that of p − n − p p-n-p p − n − p transistor?
(a) The emitter-base junction is reversed biased in n − p − n \mathrm{n}-\mathrm{p}-\mathrm{n} n − p − n transistor. (b) The emitter injects minority carriers into the base region of the p − n − p \mathrm{p}-\mathrm{n}-\mathrm{p} p − n − p transistor. (c) The emitter injects holes into the base of the p − n − p \mathrm{p}-\mathrm{n}-\mathrm{p} p − n − p transistor and electrons into the base region of n − p − n n-p-n n − p − n transistor. (d) The emitter injects holes into the base of n − p − n \mathrm{n}-\mathrm{p}-\mathrm{n} n − p − n transistor. Show answer and solution
Answer: (c)
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Q47 NEET 2023 Medium Old syllabus ☆
Junction Transistor · Structure & biasing
The above figure shows the circuit symbol of a transistor. Select the correct statements given below:
(A) The transistor has two segments of p-type semiconductor separated by a segment of n-type semiconductor.
(B) The emitter is of moderate size and heavily doped.
(C) The central segment is thin and lightly doped.
(D) The emitter base junction is reverse biased in common emitter amplifier circuit.
(a) (C) and (D) (b) (A) and (D) (c) (A) and (B) (d) (B) and (C) Show answer and solution
Answer: (d)
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Q48 MHT CET 2023 · 14 MAY · SHIFT 2 Medium ☆
Junction Transistor · Current relations
In an n-p-n transistor, the collector current is 28 mA 28\text{ mA} 28 mA . If 80 % 80\% 80% of electrons reach the collector, its base current in mA \text{mA} mA is
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Answer: (d)
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Q49 JEE Main 2023 · 8 APR · SHIFT 2 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
For a given transistor amplifier circuit in CE \text{CE} CE configuration V C C = 1 V , R C = 1 k Ω , R b = 100 k Ω \mathrm{V}_{\mathrm{CC}} = 1\text{ V}, \mathrm{R}_{\mathrm{C}} = 1\text{ k}\Omega, \mathrm{R}_{\mathrm{b}} = 100\text{ k}\Omega V CC = 1 V , R C = 1 k Ω , R b = 100 k Ω and β = 100 \beta = 100 β = 100 . Value of base current I b \mathrm{I}_{\mathrm{b}} I b is
(a) I b = 100 μ A \mathrm{I}_{\mathrm{b}} = 100\ \mu\text{A} I b = 100 μ A (b) I b = 0.1 μ A \mathrm{I}_{\mathrm{b}} = 0.1\ \mu\text{A} I b = 0.1 μ A (c) I b = 1.0 μ A \mathrm{I}_{\mathrm{b}} = 1.0\ \mu\text{A} I b = 1.0 μ A (d) I b = 10 μ A \mathrm{I}_{\mathrm{b}} = 10\ \mu\text{A} I b = 10 μ A Show answer and solution
Answer: (d)
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Q50 JEE Main 2023 · 13 APR · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
From the given transfer characteristic of a transistor in CE \text{CE} CE configuration, the value of power gain of this configuration is 1 0 x 10^{x} 1 0 x , for R B = 10 k Ω \mathrm{R}_{\mathrm{B}} = 10\text{ k}\Omega R B = 10 k Ω , and R C = 1 k Ω \mathrm{R}_{\mathrm{C}} = 1\text{ k}\Omega R C = 1 k Ω . The value of x x x is __________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 3
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Q51 JEE Main 2022 · 27 JUL · SHIFT 2 Easy Old syllabus ☆
Junction Transistor · Current relations
For a constant collector-emitter voltage of 8 V 8\text{ V} 8 V , the collector current of a transistor reached to the value of 6 mA 6\text{ mA} 6 mA from 4 mA 4\text{ mA} 4 mA , whereas base current changed from 20 μ A 20\,\mu\text{A} 20 μ A to 25 μ A 25\,\mu\text{A} 25 μ A value. If transistor is in active state, small signal current gain (current amplification factor) will be:
(a) 240 (b) 400 (c) 0.0025 (d) 200 Show answer and solution
Answer: (b)
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Q52 JEE Main 2022 · 27 JUN · SHIFT 2 Easy Old syllabus ☆
Junction Transistor · Switch & oscillator
For a transistor to act as a switch, it must be operated in
(a) Active region. (b) Saturation state only. (c) Cut-off state only. (d) Saturation and cut-off state. Show answer and solution
Answer: (d)
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Q53 JEE Main 2022 · 26 JUN · SHIFT 2 Easy Old syllabus ☆
Junction Transistor · Switch & oscillator
The positive feedback is required by an amplifier to act an oscillator. The feedback here means:
(a) External input is necessary to sustain ac signal in output. (b) A portion of the output power is returned back to the input. (c) Feedback can be achieved by LR network. (d) The base-collector junction must be forward biased. Show answer and solution
Answer: (b)
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Q54 JEE Main 2022 · 28 JUN · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · Structure & biasing
Given below are two statements: One is labelled as Assertion A and the other is labelled as Reason R.
Assertion A: n-p-n transistor permits more current than a p-n-p transistor.
Reason R: Electrons have greater mobility as a charge carrier.
Choose the correct answer from the options given below:
(a) Both A and R are true, and R is correct explanation of A. (b) Both A and R are true but R is NOT the correct explanation of A. (c) A is true but R is false. (d) A is false but R is true. Show answer and solution
Answer: (a)
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Q55 NEET 2022 · PHASE 2 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
The collector current in a common base amplifier using n-p-n transistor is 24 mA. If 80% of the electrons released by the emitter is accepted by the collector, then the base current is numerically:
(a) 3 mA and entering the base (b) 6 mA and leaving the base (c) 3 mA and leaving the base (d) 6 mA and entering the base Show answer and solution
Answer: (d)
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Q56 JEE Main 2022 · 28 JUL · SHIFT 2 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
An n.p.n transistor with current gain β = 100 \beta = 100 β = 100 in common emitter configuration is shown in figure. The output voltage of the amplifier will be
(a) 0.1 V (b) 1.0 V (c) 10 V (d) 100 V Show answer and solution
Answer: (b)
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Q57 JEE Main 2022 · 29 JUN · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
A transistor is used in an amplifier circuit in common emitter mode. If the base current changes by 100 μ \mu μ A, it brings a change of 10 mA in collector current. If the load resistance is 2 kΩ \Omega Ω and input resistance is 1 kΩ \Omega Ω , the value of power gain is x × \times × 104. The value of x is _____________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 2
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Q58 JEE Main 2022 · 24 JUN · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
A transistor is used in common-emitter mode in an amplifier circuit. When a signal of 10 mV is added to the base-emitter voltage, the base current changes by 10 μ \mu μ A and the collector current changes by 1.5 mA. The load resistance is 5 kΩ \Omega Ω . The voltage gain of the transistor will be _________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 750
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Q59 JEE Main 2022 · 26 JUL · SHIFT 2 Medium Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
The typical transfer characteristics of a transistor in CE configuration is shown in figure. A load resistor of 2 k Ω 2\text{ k}\Omega 2 k Ω is connected in the collector branch of the circuit used. The input resistance of the transistor is 0.50 k Ω 0.50\text{ k}\Omega 0.50 k Ω . The voltage gain of the transistor is ______________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 200
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Q60 JEE Main 2022 · 25 JUN · SHIFT 2 Medium Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
In an experiment of CE configuration of n-p-n transistor, the transfer characteristics are observed as given in figure.
If the input resistance is 200 Ω \Omega Ω and output resistance is 60 Ω \Omega Ω , the voltage gain in this experiment will be ____________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 15
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Q61 MHT CET 2021 · 24 SEP · SHIFT 2 Easy ☆
Junction Transistor · Current relations
For a transistor, the current ratio α dc = 69 70 \alpha_\text{dc} = \frac{69}{70} α dc = 70 69 , the current gain β dc \beta_\text{dc} β dc is
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Answer: (b)
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Q62 MHT CET 2021 · 21 SEP · SHIFT 1 Easy ☆
Junction Transistor · Current relations
For a transistor, 1 α DC − 1 β DC \frac{1}{\alpha_\text{DC}} - \frac{1}{\beta_\text{DC}} α DC 1 − β DC 1 is equal to [ α DC \alpha_\text{DC} α DC and β DC \beta_\text{DC} β DC are current amplification factors]
(a) three (b) two (c) zero (d) one Show answer and solution
Answer: (d)
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Q63 JEE Main 2021 · 27 AUG · SHIFT 2 Easy Old syllabus ☆
Junction Transistor · Current relations
For a transistor α \alpha α and β \beta β are given as α = I C I E \alpha = \frac{I_C}{I_E} α = I E I C and β = I C I B \beta = \frac{I_C}{I_B} β = I B I C . Then the correct relation between α \alpha α and β \beta β will be:
(a) α = 1 − β β \alpha = \frac{1-\beta}{\beta} α = β 1 − β (b) β = α 1 − α \beta = \frac{\alpha}{1-\alpha} β = 1 − α α (c) α β = 1 \alpha\beta = 1 α β = 1 (d) α = β 1 − β \alpha = \frac{\beta}{1-\beta} α = 1 − β β Show answer and solution
Answer: (b)
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Q64 JEE Main 2021 · 18 MAR · SHIFT 2 Easy Old syllabus ☆
Junction Transistor · Current relations
The correct relation between α \alpha α (ratio of collector current to emitter current) and β \beta β (ratio of collector current to base current) of a transistor is:
(a) β = α 1 + α \beta = \frac{\alpha}{1+\alpha} β = 1 + α α (b) α = β 1 − α \alpha = \frac{\beta}{1-\alpha} α = 1 − α β (c) α = β 1 + β \alpha = \frac{\beta}{1+\beta} α = 1 + β β (d) β = 1 1 − α \beta = \frac{1}{1-\alpha} β = 1 − α 1 Show answer and solution
Answer: (c)
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Q65 JEE Main 2021 · 24 FEB · SHIFT 1 Easy Old syllabus ☆
Junction Transistor · Current relations
If an emitter current is changed by 4 mA, the collector current changes by 3.5 mA. The value of β \beta β will be:
Show answer and solution
Answer: (d)
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Q66 MHT CET 2021 · 22 SEP · SHIFT 2 Easy ☆
Junction Transistor · CE amplifier gains
For a common-emitter amplifier, the voltage gain is 40. Its input and output impedances are 100 Ω 100\ \Omega 100 Ω and 400 Ω 400\ \Omega 400 Ω , respectively. The power gain of the CE amplifier will be
Show answer and solution
Answer: (b)
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Q67 JEE Main 2021 · 27 AUG · SHIFT 1 Easy Old syllabus ☆
Junction Transistor · CE amplifier gains
For a transistor in CE mode to be used as an amplifier, it must be operated in:
(a) Both cut-off and Saturation (b) Saturation region only (c) Cut-off region only (d) The active region only Show answer and solution
Answer: (d)
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Q68 JEE Main 2021 · 24 FEB · SHIFT 2 Medium Old syllabus ☆
Junction Transistor · Structure & biasing
Given below are two statements:
Statement I: PN junction diodes can be used to function as transistor, simply by connecting two diodes, back to back, which acts as the base terminal.
Statement II: In the study of transistor, the amplification factor β \beta β indicates ratio of the collector current to the base current.
In the light of the above statements, choose the correct answer from the options given below.
(a) Both Statement I and Statement II are false (b) Statement I is false but Statement II is true (c) Both Statement I and Statement II are true (d) Statement I is true but Statement II is false Show answer and solution
Answer: (b)
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Q69 MHT CET 2021 · 23 SEP · SHIFT 2 Medium ☆
Junction Transistor · Current relations
In a CE transistor, a change of 8.0 mA 8.0\text{ mA} 8.0 mA in the emitter current produces a change of 7.8 mA 7.8\text{ mA} 7.8 mA in the collector current. What change in the base current is necessary to produce the same change in the collector current?
(a) 200 μ A 200\ \mu\text{A} 200 μ A (b) 50 μ A 50\ \mu\text{A} 50 μ A (c) 100 μ A 100\ \mu\text{A} 100 μ A (d) 150 μ A 150\ \mu\text{A} 150 μ A Show answer and solution
Answer: (a)
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Q70 JEE Main 2021 · 27 JUL · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · Current relations
A transistor is connected in common emitter circuit configuration, the collector supply voltage is 10 V and the voltage drop across a resistor of 1000 Ω \Omega Ω in the collector circuit is 0.6 V. If the current gain factor (β \beta β ) is 24, then the base current is _____________ μ \mu μ A. (Round off to the Nearest Integer)
Numerical answer type. Work it out, then open the answer.
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Answer: 25
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Q71 MHT CET 2021 · 23 SEP · SHIFT 1 Medium ☆
Junction Transistor · CE amplifier gains
In common emitter amplifier, a change of 0.2 mA in the base current causes a change of 5 mA in the collector current. If input resistance is 2 kΩ \Omega Ω and voltage gain is 75, the load resistance used in the circuit is
(a) 8 kΩ \Omega Ω (b) 4 kΩ \Omega Ω (c) 12 kΩ \Omega Ω (d) 6 kΩ \Omega Ω Show answer and solution
Answer: (d)
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Q72 JEE Main 2021 · 18 MAR · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
An npn transistor operates as a common emitter amplifier with a power gain of 106. The input circuit resistance is 100Ω \Omega Ω and the output load resistance is 10 KΩ \Omega Ω . The common emitter current gain 'β \beta β ' will be ________. (Round off to the Nearest Integer).
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 100
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Q73 JEE Main 2021 · 18 MAR · SHIFT 2 Medium Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
The typical output characteristics curve for a transistor working in the common-emitter configuration is shown in the figure.
The estimated current gain from the figure is __________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 200
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Q74 MHT CET 2021 · 22 SEP · SHIFT 1 Difficult ☆
Junction Transistor · Current relations
In an n-p-n transistor 200 electrons enter the emitter in 10 − 8 ^{-8} − 8 second. If 1% electrons are lost in the base, then the current that enters the emitter and the current amplification factor are respectively [e = 1.6 × \times × 10 − 19 ^{-19} − 19 C]
(a) 2 × \times × 10 − 10 ^{-10} − 10 A and 49 (b) 3.2 × \times × 10− 9 ^{-9} − 9 A and 99 (c) 1.6 × \times × 10− 19 ^{-19} − 19 A and 90 (d) 1.7 × \times × 10 − 11 ^{-11} − 11 A and 70 Show answer and solution
Answer: (b)
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Q75 JEE Main 2021 · 27 AUG · SHIFT 1 Difficult Old syllabus ☆
Junction Transistor · CE amplifier gains
A circuit is arranged as shown in figure. The output voltage V0 is equal to ................... V.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 5
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Q76 NEET 2020 · PHASE 1 Easy Old syllabus ☆
Junction Transistor · Structure & biasing
For transistor action, which of the following statements is correct?
(a) Base, emitter and collector regions should have same size. (b) Both emitter junction as well as the collector junction are forwarded biased. (c) The base region must be very thin and lightly doped. (d) Base, emitter and collector regions should have same doping concentrations. Show answer and solution
Answer: (c)
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Q77 MHT CET 2020 · 19 OCT · SHIFT 2 Easy ☆
Junction Transistor · Current relations
A transistor having α = 0.8 \alpha = 0.8 α = 0.8 is connected in common emitter configuration. When the base current changes by 6 mA, then the change in collector current is
(a) 12 mA (b) 24 mA (c) 0.66 mA (d) 1.5 mA Show answer and solution
Answer: (b)
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Q78 MHT CET 2020 · 16 OCT · SHIFT 1 Medium ☆
Junction Transistor · CE amplifier gains
In common emitter amplifier, input resistance is 1000 Ω 1000\ \Omega 1000 Ω , peak value of input signal voltage is 5 mV 5\text{ mV} 5 mV and β = 60 \beta = 60 β = 60 . The peak value of output current is
(a) 3 × 1 0 − 4 A 3 \times 10^{-4}\text{ A} 3 × 1 0 − 4 A (b) 0.5 × 1 0 − 4 A 0.5 \times 10^{-4}\text{ A} 0.5 × 1 0 − 4 A (c) 2 × 1 0 − 5 A 2 \times 10^{-5}\text{ A} 2 × 1 0 − 5 A (d) 1 × 1 0 − 5 A 1 \times 10^{-5}\text{ A} 1 × 1 0 − 5 A Show answer and solution
Answer: (a)
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Q79 NEET 2020 · PHASE 2 Medium ☆
Junction Transistor · CE amplifier gains
A n-p-n transistor is connected in common emitter configuration (see figure) in which collector voltage drop across load resistance ( 800 Ω ) (800\ \Omega) ( 800 Ω ) connected to the collector circuit is 0.8 V 0.8\text{ V} 0.8 V . The collector current is
(a) 0.2 mA 0.2\text{ mA} 0.2 mA (b) 2 mA 2\text{ mA} 2 mA (c) 0.1 mA 0.1\text{ mA} 0.1 mA (d) 1 mA 1\text{ mA} 1 mA Show answer and solution
Answer: (d)
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Q80 JEE Main 2020 · 6 SEP · SHIFT 2 Medium Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
The output characteristics of a transistor is shown in the figure.
When VCE is 10V and IC = 4.0 mA, then value of β \beta β ac is __________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 150
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Q81 MHT CET 2020 · 16 OCT · SHIFT 2 Medium ☆
Junction Transistor · Switch & oscillator
A A A transistor has a voltage gain A A A . If the amount β A \beta A β A of its output is applied to the input of the transistor, then the transistor becomes oscillator, when
(a) β A = 0 \beta A = 0 β A = 0 (b) β A = 1 \beta A = 1 β A = 1 (c) β A = ∞ \beta A = \infty β A = ∞ (d) β = 0 \beta = 0 β = 0 Show answer and solution
Answer: (b)
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Q82 MHT CET 2019 · 3 MAY · SHIFT 1 Easy ☆
Junction Transistor · Structure & biasing
Which of the following regions of a transistors are, respectively, heavily dopped and lightly dopped?
(a) Collector and emitter (b) Base and emitter (c) Emitter and base (d) Emitter and collector Show answer and solution
Answer: (c)
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Q83 MHT CET 2019 · 2 MAY · SHIFT 2 Easy ☆
Junction Transistor · Current relations
In the study of transistor as an amplifier, the ratio of collector current to emitter current is 0.98 then the ratio of collector current to base current will be
Show answer and solution
Answer: (b)
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Q84 MHT CET 2019 · 2 MAY · SHIFT 1 Easy ☆
Junction Transistor · Current relations
. For a transistor, the current ratio ' β d c \beta_{dc} β d c ' is defined as the ratio of
(a) collector current to emitter current. (b) collector current to base current. (c) base current to collector current. (d) emitter current to collector current. Show answer and solution
Answer: (b)
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Q85 JEE Main 2019 · 10 APR · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
An n-p-n transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is 100Ω \Omega Ω and the output load resistance is 10 kΩ \Omega Ω . The common emitter current gain β \beta β is:
(a) 104 (b) 102 (c) 6 × 102 (d) 60 Show answer and solution
Answer: (b)
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Q86 JEE Main 2019 · 9 APR · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
An NPN transistor is used in common emitter configuration as an amplifier with 1 kΩ \Omega Ω load resistance. Signal voltage of 10 mV is applied across the base-emitter. This produces a 3 mA change in the collector current and 15μA change in the base current of the amplifier. The input resistance and voltage gain are:
(a) 0.67 kΩ \Omega Ω , 200 (b) 0.33 kΩ \Omega Ω , 1.5 (c) 0.67 kΩ \Omega Ω , 300 (d) 0.33 kΩ \Omega Ω , 300 Show answer and solution
Answer: (c)
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Q87 JEE Main 2019 · 12 APR · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
The transfer characteristic curve of a transistor, having input and output resistance 100 Ω \Omega Ω and 100 kΩ \Omega Ω respectively is shown in the figure. The voltage and power gain, are respectively:
(a) 5 × 104, 5 × 105 (b) 5 × 104, 5 × 106 (c) 5 × 104, 2.5 × 106 (d) 2.5 × 104, 2.5 × 106 Show answer and solution
Answer: (c)
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Q88 JEE Main 2019 · 12 JAN · SHIFT 2 Difficult Old syllabus ☆
Junction Transistor · Current relations
In the figure, given that VBB supply can vary from 0 to 5.0 V, VCC = 5V, β \beta β dc = 200, RB = 100 kΩ \Omega Ω , RC = 1 kΩ \Omega Ω and VBE = 1.0 V. The minimum base current and the input voltage at which the transistor will go to saturation, will be respectively:
(a) 20 μ \mu μ A and 2.8 V (b) 25 μ \mu μ A and 2.8 V (c) 20 μ \mu μ A and 3.5 V (d) 25 μ \mu μ A and 3.5 V Show answer and solution
Answer: (d)
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Q89 JEE Main 2019 · 8 APR · SHIFT 2 Difficult Old syllabus ☆
Junction Transistor · CE amplifier gains
A common emitter amplifier circuit, built using an npn transistor, is shown in the figure. Its dc current gain is 250, RC = 1kΩ \Omega Ω and VCC = 10 V. What is the minimum base current for VCE to reach saturation?
(a) 100 μ \mu μ A (b) 7 μ \mu μ A (c) 10 μ \mu μ A (d) 40 μ \mu μ A Show answer and solution
Answer: (d)
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Q90 JEE Main 2018 · 15 APR · SHIFT 1 Easy Old syllabus ☆
Junction Transistor · CE amplifier gains
In a common emitter configuration with suitable bias, it is given that R L R_L R L is the load resistance and R B E R_{BE} R BE is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by:
β \beta β is current gain, I B , I C \mathrm{I}_B, \mathrm{I}_C I B , I C and I E \mathrm{I}_E I E are respectively base, collector and emitter currents.
(a) β R L R B E , Δ I C Δ I B , β 2 R L R B E \beta\frac{R_L}{R_{BE}}, \frac{\Delta \mathrm{I}_C}{\Delta \mathrm{I}_B}, \beta^2\frac{R_L}{R_{BE}} β R BE R L , Δ I B Δ I C , β 2 R BE R L (b) β R L R B E , Δ I E Δ I B , β 2 R L R B E \beta\frac{R_L}{R_{BE}}, \frac{\Delta \mathrm{I}_E}{\Delta \mathrm{I}_B}, \beta^2\frac{R_L}{R_{BE}} β R BE R L , Δ I B Δ I E , β 2 R BE R L (c) β 2 R L R B E , Δ I C Δ I E , β 2 R L R B E \beta^2\frac{R_L}{R_{BE}}, \frac{\Delta \mathrm{I}_C}{\Delta \mathrm{I}_E}, \beta^2\frac{R_L}{R_{BE}} β 2 R BE R L , Δ I E Δ I C , β 2 R BE R L (d) β 2 R L R B E , Δ I C Δ I B , β R L R B E \beta^2\frac{R_L}{R_{BE}}, \frac{\Delta \mathrm{I}_C}{\Delta \mathrm{I}_B}, \beta\frac{R_L}{R_{BE}} β 2 R BE R L , Δ I B Δ I C , β R BE R L Show answer and solution
Answer: (a)
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Q91 NEET 2018 Medium Old syllabus ☆
Junction Transistor · Current relations
In the circuit shown in the figure, the input voltage Vi is 20 V, VBE = 0 and VCE = 0. The values of IB, IC and β \beta β are given by
(a) IB = 40 μ \mu μ A, IC = 10 mA, β \beta β = 250 (b) IB = 25 μ \mu μ A, IC = 5 mA, β \beta β = 200 (c) IB = 20 μ \mu μ A, IC = 5 mA, β \beta β = 250 (d) IB = 40 μ \mu μ A, IC = 5 mA, β \beta β = 125 Show answer and solution
Answer: (d)
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Q92 JEE Main 2017 Easy Old syllabus ☆
Junction Transistor · CE amplifier gains
In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be:
(a) 180° (b) 45° (c) 90° (d) 135° Show answer and solution
Answer: (a)
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Q93 JEE Main 2017 · 9 APR · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · Current relations
The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is:
(a) 9.6 mA (b) 6.9 mA (c) 0.69 mA (d) 69 mA Show answer and solution
Answer: (b)
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Q94 NEET 2017 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
In a common emitter transistor amplifier the audio sgnal voltage across the collector is 3 V. The resistance of collector is 3 kΩ \Omega Ω . If current gain is 100 and the base resistance is 2 kΩ \Omega Ω , the voltage and power gain of the amplifier is
(a) 15 and 200 (b) 150 and 15000 (c) 20 and 2000 (d) 200 and 1000 Show answer and solution
Answer: (b)
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Q95 JEE Main 2016 Easy Old syllabus ☆
Junction Transistor · Current relations
For a common emitter configuration, if α \alpha α and β \beta β have their usual meanings, the incorrect relationship between α \alpha α and β \beta β is:
(a) α = β 1 + β \alpha = \frac{\beta}{1+\beta} α = 1 + β β (b) α = β 2 1 + β 2 \alpha = \frac{\beta^2}{1+\beta^2} α = 1 + β 2 β 2 (c) 1 α = 1 β + 1 \frac{1}{\alpha} = \frac{1}{\beta} + 1 α 1 = β 1 + 1 (d) α = β 1 − β 2 \alpha = \frac{\beta}{1-\beta^2} α = 1 − β 2 β Show answer and solution
Answer: (b)
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Q96 NEET 2016 · PHASE 2 Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
For CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ \Omega Ω is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ \Omega Ω , then the input signal voltage is
(a) 10 mV (b) 20 mV (c) 30 mV (d) 15 mV Show answer and solution
Answer: (b)
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Q97 JEE Main 2016 · 10 APR · SHIFT 1 Medium Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
The ratio (R) of output resistance r0, and the input resistance ri in measurements of input and output characteristics of a transistor is typically in the range:
(a) R ~ 102 − - − 103 (b) R ~ 1 − - − 10 (c) R ~ 0.1 − - − 0.01 (d) R ~ 0.1 − - − 1.0 Show answer and solution
Answer: (b)
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Q98 NEET 2016 · PHASE 1 Difficult Old syllabus ☆
Junction Transistor · CE amplifier gains
A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω \Omega Ω is connected in the collector circuit and the voltage drop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω \Omega Ω , the voltage gain and the power gain of the amplifier will respectively be
(a) 4, 4 (b) 4, 3.69 (c) 4, 3.84 (d) 3.69, 3.84 Show answer and solution
Answer: (c)
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Q99 JEE Main 2016 · 10 APR · SHIFT 1 Difficult Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
A realistic graph depicting the variation of the reciprocal of input resistance in an input characteristics measurement in a common emitter transistor configuration is:
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Answer: (c)
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Q100 JEE Main 2016 · 9 APR · SHIFT 1 Difficult Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
An unknown transistor needs to be identified as a npn or pnp type. A multimeter, with + ve and − ve terminals, is used to measure resistance between different terminals of transistor. If terminal 2 is the base of the transistor then which of the following is correct for a pnp transistor?
(a) + ve terminal 1, − - − ve terminal 2, resistance high (b) + ve terminal 2, − - − ve terminal 1, resistance high (c) + ve terminal 3, − - − ve terminal 2, resistance high (d) + ve terminal 2, − - − ve terminal 3, resistance low Show answer and solution
Answer: (a)
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Q101 NEET 2015 · AIPMT Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
The input signal given to a CE amplifier having a voltage gain of 150 is Vi = 2cos(15t + π 3 \frac{\pi}{3} 3 π ). The corresponding output signal will be
(a) 2cos ( 15 t + 5 π 6 ) \left(15t + \frac{5\pi}{6}\right) ( 15 t + 6 5 π ) (b) 300cos ( 15 t + 4 π 3 ) \left(15t + \frac{4\pi}{3}\right) ( 15 t + 3 4 π ) (c) 300cos ( 15 t + π 3 ) \left(15t + \frac{\pi}{3}\right) ( 15 t + 3 π ) (d) 75cos ( 15 t + 2 π 3 ) \left(15t + \frac{2\pi}{3}\right) ( 15 t + 3 2 π ) Show answer and solution
Answer: (b)
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Q102 JEE Main 2014 · 9 APR · SHIFT 1 Difficult Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
An n-p-n transistor has three leads A, B and C. Connecting B and C by moist fingers, A to the positive lead of an ammeter, and C to the negative lead of the ammeter, one finds large deflection. Then, A, B and C refer respectively to:
(a) Emitter, base and collector (b) Base, emitter and collector (c) Base, collector and emitter (d) Collector, emitter and base. Show answer and solution
Answer: (a)
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Q103 NEET 2013 Medium Old syllabus ☆
Junction Transistor · Structure & biasing
One way in which the operation of a n-p-n transistor differs from that of a p-n-p
(a) The emitter junction injects minority carriers into the base region of the p-n-p (b) The emitter injects holes into the base of the p-n-p and electrons into the base region of n-p-n (c) The emitter injects holes into the base of n-p-n (d) The emitter junction is reverse biased in n-p-n Show answer and solution
Answer: (b)
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Q104 NEET 2013 Difficult Old syllabus ☆
Junction Transistor · CE amplifier gains
In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be
(a) 1 3 G \frac{1}{3}G 3 1 G (b) 5 4 G \frac{5}{4}G 4 5 G (c) 2 3 G \frac{2}{3}G 3 2 G (d) 1.5G Show answer and solution
Answer: (c)
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Q105 NEET 2012 · AIPMT · PRELIMS Easy Old syllabus ☆
Junction Transistor · Switch & oscillator
Transfer characteristics [output voltage (V0) vs input voltage (Vi)] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used
(a) in region III (b) both in region (I) and (III) (c) in region II (d) in region I Show answer and solution
Answer: (b)
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Q106 NEET 2012 · AIPMT · MAINS Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
The input resistance of a silicon transistor is 100 Ω \Omega Ω . Base current is changed by 40 μ \mu μ A which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 kΩ \Omega Ω . The voltage gain of the amplifier is
(a) 2000 (b) 3000 (c) 4000 (d) 1000 Show answer and solution
Answer: (a)
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Q107 NEET 2012 · AIPMT · PRELIMS Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 kΩ \Omega Ω is 2 V. If the base resistance is 1 kΩ \Omega Ω and the current amplification of the transistor is 100, the input signal voltage is
(a) 0.1 V (b) 1.0 V (c) 1 mV (d) 10 mV Show answer and solution
Answer: (d)
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Q108 NEET 2011 · AIPMT · PRELIMS Easy Old syllabus ☆
Junction Transistor · Current relations
A transistor is operated in common emitter configuration at VC = 2 V such that a change in the base current from 100 μ \mu μ A to 300 μ \mu μ A produces a change in the collector current from 10 mA to 20 mA. The current gain is
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Answer: (a)
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Q109 NEET 2010 · AIPMT · MAINS Easy Old syllabus ☆
Junction Transistor · Structure & biasing
For transistor action
(1) Base, emitter and collector regions should have similar size and doping concentrations. (2) The base region must be very thin and lightly doped.
(3) The emitter-base junction is forward biased and base-collector junction is reverse biased. (4) Both the emitter-base junction as well as the base-collector junction are forward biased. Which one of the following pairs of statements is correct?
(a) (4) and (1) (b) (1) and (2) (c) (2) and (3) (d) (3) and (4) Show answer and solution
Answer: (c)
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Q110 NEET 2010 · AIPMT · PRELIMS Easy Old syllabus ☆
Junction Transistor · CE amplifier gains
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω \Omega Ω and an output impedance of 200 Ω \Omega Ω . The power gain of the amplifier is
(a) 500 (b) 1000 (c) 1250 (d) 50 Show answer and solution
Answer: (c)
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Q111 NEET 2009 · AIPMT Easy Old syllabus ☆
Junction Transistor · Current relations
A transistor is operated in common-emitter configuration at VC = 2V such that a change in the base current from 100 μ \mu μ A to 200 μ \mu μ A produces a change in the collector current from 5 mA to 10 mA. The current gain is
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Answer: (c)
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Q112 NEET 2008 · AIPMT Medium Old syllabus ☆
Junction Transistor · Switch & oscillator
The voltage gain of an amplifier with 9% negative feedback is 10. The voltage gain without feedback will be
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Answer: (b)
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Q113 JEE Main 2008 · AIEEE Difficult Old syllabus ☆
Junction Transistor · Characteristics & transistor testing
A working transistor with its three legs marked P , Q P, Q P , Q and R R R is tested using a multi-meter. No conduction is found between P P P and Q Q Q . By connecting the common (negative) terminal of the multi- meter to R R R and the other (positive) terminal to P P P or Q , Q, Q , some resistance is seen on the multi-meter. Which of the following is true for the transistor?
(a) It is an n p n npn n p n transistor with R R R as base (b) It is an p n p pnp p n p transistor with R R R as collector (c) It is an p n p pnp p n p transistor with R R R as emitter (d) It is an n p n npn n p n transistor with R R R as collector Show answer and solution
Answer: (a)
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Q114 NEET 2007 · AIPMT Easy ☆
Junction Transistor · CE amplifier gains
A common emitter amplifier has a voltage gain of 50, an input impedance of 100 Ω 100\ \Omega 100 Ω and an output impedance of 200 Ω 200\ \Omega 200 Ω . The power gain of the amplifier is
(a) 1000 1000 1000 (b) 1250 1250 1250 (c) 100 100 100 (d) 500 500 500 Show answer and solution
Answer: (b)
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Q115 NEET 2006 · AIPMT Easy Old syllabus ☆
Junction Transistor · Current relations
A transistor is operated in common emitter configuration at constant collector voltage VC = 1.5 V such that a change in the base current from 100 μ \mu μ A to 150 μ \mu μ A produces a change in the collector current from 5 mA to 10 mA. The current gain β \beta β is
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Answer: (d)
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Q116 JEE Main 2006 · AIEEE Easy Old syllabus ☆
Junction Transistor · Current relations
In a common base mode of a transistor, the collector current is 5.488 m A 5.488\,mA 5.488 m A for an emitter current of 5.60 m A . 5.60\,mA. 5.60 m A . The value of the base current amplification factor ( β ) (\beta) ( β ) will be
(a) 49 49 49 (b) 50 50 50 (c) 51 51 51 (d) 48 48 48 Show answer and solution
Answer: (a)
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Q117 JEE Main 2005 · AIEEE Easy Old syllabus ☆
Junction Transistor · CE amplifier gains
In a common base amplifier, the phase difference between the input signal voltage and output voltage is
(a) π \pi π (b) π 4 \frac{\pi}{4} 4 π (c) π 2 \frac{\pi}{2} 2 π (d) 0 0 0 Show answer and solution
Answer: (d)
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Q118 JEE Main 2004 · AIEEE Easy Old syllabus ☆
Junction Transistor · Structure & biasing
When n p n npn n p n transistor is used as an amplifer
(a) electrons move from collector to base (b) holes move from emitter to base (c) electrons move from base to collector (d) holes move from base to emitter Show answer and solution
Answer: (c)
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Q119 JEE Main 2004 · AIEEE Medium Old syllabus ☆
Junction Transistor · CE amplifier gains
For a transistor amplifier in common emitter configuration for load impedance of 1 k Ω ( h f e = 50 1\text{ k}\Omega\ (h_{fe} = 50 1 k Ω ( h f e = 50 and h o e = 25 μ A/V ) h_{oe} = 25\ \mu\text{A/V}) h oe = 25 μ A/V ) the current gain is
(a) − 24.8 -24.8 − 24.8 (b) − 15.7 -15.7 − 15.7 (c) − 5.2 -5.2 − 5.2 (d) − 48.78 -48.78 − 48.78 Show answer and solution
Answer: (d)
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Q120 NEET 2003 · AIPMT Easy Old syllabus ☆
Junction Transistor · Structure & biasing
A n-p-n transistor conducts when
(a) both collector and emitter are positive with respect to the base (b) collector is positive and emitter is negative with respect to the base (c) collector is positive and emitter is at same potential as the base (d) both collector and emitter are negative with respect to the base Show answer and solution
Answer: (b)
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Q121 JEE Main 2002 · AIEEE Easy Old syllabus ☆
Junction Transistor · Structure & biasing
The part of a transistor which is most heavily doped to produce large number of majority carriers is
(a) emmiter (b) base (c) collector (d) can be any of the above three Show answer and solution
Answer: (a)
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Q122 NEET 2002 · AIPMT Easy Old syllabus ☆
Junction Transistor · Current relations
For a transistor I C I E \frac{I_C}{I_E} I E I C = 0.96, then current gain for common emitter is
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Answer: (d)
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Q123 NEET 2001 · AIPMT Easy Old syllabus ☆
Junction Transistor · Current relations
For a common base circuit if I C I E \frac{I_C}{I_E} I E I C = 0.98 then current gain for common emitter circuit will be
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Answer: (a)
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Q124 NEET 2000 · AIPMT Easy Old syllabus ☆
Junction Transistor · Current relations
The correct relation for α , β \alpha, \beta α , β for a transistor
(a) β = 1 − α α \beta = \frac{1-\alpha}{\alpha} β = α 1 − α (b) β = α 1 − α \beta = \frac{\alpha}{1-\alpha} β = 1 − α α (c) α = β − 1 β \alpha = \frac{\beta-1}{\beta} α = β β − 1 (d) α β = 1 \alpha\beta = 1 α β = 1 Show answer and solution
Answer: (b)
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Q125 NEET 1998 · AIPMT Medium ☆
Junction Transistor · CE amplifier gains
The transfer ratio β \beta β of a transistor is 50 50 50 . The input resistance of the transistor when used in the common emitter configuration is 1 k Ω 1\text{ k}\Omega 1 k Ω . The peak value of the collector AC current for an AC input voltage of 0.01 V 0.01\text{ V} 0.01 V peak is
(a) 100 μ A 100\ \mu\text{A} 100 μ A (b) 0.01 mA 0.01\text{ mA} 0.01 mA (c) 0.25 mA 0.25\text{ mA} 0.25 mA (d) 500 μ A 500\ \mu\text{A} 500 μ A Show answer and solution
Answer: (d)
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Q126 NEET 1997 · AIPMT Easy ☆
Junction Transistor · Current relations
The correct relationship between the two current gains α \alpha α and β \beta β in a transistor is
(a) α = β 1 + β \alpha = \frac{\beta}{1 + \beta} α = 1 + β β (b) α = 1 + β β \alpha = \frac{1 + \beta}{\beta} α = β 1 + β (c) β = α 1 + α \beta = \frac{\alpha}{1 + \alpha} β = 1 + α α (d) β = α α − 1 \beta = \frac{\alpha}{\alpha - 1} β = α − 1 α Show answer and solution
Answer: (a)
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Q127 NEET 1996 · AIPMT Easy ☆
Junction Transistor · CE amplifier gains
When npn transistor is used as an amplifier, then
(a) electrons move from collector to base (b) holes move from base to emitter (c) electrons move from base to collector (d) electrons move from emitter to base Show answer and solution
Answer: (c)
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Q128 NEET 1993 · AIPMT Easy ☆
Junction Transistor · Structure & biasing
The part of the transistor which is heavily doped to produce large number of majority carriers is
(a) emitter (b) base (c) collector (d) any of the above depending upon the nature of transistor Show answer and solution
Answer: (a)
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Q129 NEET 1991 · AIPMT Easy ☆
Junction Transistor · CE amplifier gains
To use a transistor as an amplifier
(a) the emitter base junction is forward biased and the base collector junction is reversed biased (b) no bias voltage is required (c) both junction are forward biased (d) both junction are reverse biased Show answer and solution
Answer: (a)
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Q130 NEET 1990 · AIPMT Easy ☆
Junction Transistor · CE amplifier gains
In a common base amplifier the phase difference between the input signal voltage and the output voltage is
(a) 0 0 0 (b) π 4 \frac{\pi}{4} 4 π (c) π 2 \frac{\pi}{2} 2 π (d) π \pi π Show answer and solution
Answer: (a)
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Q131 NEET 1989 · AIPMT Easy Old syllabus ☆
Junction Transistor · Switch & oscillator
Radiowaves of constant amplitude can be generated with
(a) FET (b) filter (c) rectifier (d) oscillator Show answer and solution
Answer: (d)
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