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 are given by (a) IB = 40 μ A, IC = 10 mA, β = 250 (b) IB = 25 μ A, IC = 5 mA, β = 200 (c) IB = 20 μ A, IC = 5 mA, β = 250 (d) IB = 40 μ A, IC = 5 mA, β = 125

Answer: (d). Base loop with VBE = 0: IB = (Vi)/(RB) = (20)/(500 × 10³) = 40 μ A. Collector loop with VCE = 0: IC = (20)/(4 × 10³) = 5 mA. β = (IC)/(IB) = 5 × 10⁻³40 × 10⁻⁶ = 125 So the answer is (d).

In the circuit shown in the figure, the input voltage Vi is 20 V, VBE… | NEET 2018

Q91NEET 2018Medium

Junction Transistor

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 $\beta$ are given by

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Answer: (d)

Base loop with $V_{BE} = 0$ : $I_B = \dfrac{V_i}{R_B} = \dfrac{20}{500 \times 10^3} = 40\ \mu$ A.
Collector loop with $V_{CE} = 0$ : $I_C = \dfrac{20}{4 \times 10^3} = 5$ mA.
$\beta = \dfrac{I_C}{I_B} = \dfrac{5 \times 10^{-3}}{40 \times 10^{-6}} = 125$
So the answer is (d).

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