Q1 MHT CET 2026 · 19 MAY · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating voltage is given by V = V 0 sin ( ω t + π 4 ) V = V_0 \sin\left(\omega t + \frac{\pi}{4}\right) V = V 0 sin ( ω t + 4 π )
(T T T cos 0 ∘ = sin 9 0 ∘ = 1 \cos 0^\circ = \sin 90^\circ = 1 cos 0 ∘ = sin 9 0 ∘ = 1
(a) T 4 \frac{T}{4} 4 T (b) T 6 \frac{T}{6} 6 T (c) T 8 \frac{T}{8} 8 T (d) T 12 \frac{T}{12} 12 T Show answer and solution
Answer: (c)
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Report a mistake Q2 MHT CET 2026 · 15 MAY · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An AC source is 120 120 120 60 60 60 1 720 \frac{1}{720} 720 1 ( sin 3 0 ∘ = 1 2 ) \left(\sin 30^\circ = \frac{1}{2}\right) ( sin 3 0 ∘ = 2 1 )
(a) 20.2 V 20.2\text{ V} 20.2 V (b) 42.4 V 42.4\text{ V} 42.4 V (c) 84.8 V 84.8\text{ V} 84.8 V (d) 106.9 V 106.9\text{ V} 106.9 V Show answer and solution
Answer: (c)
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Report a mistake Q3 MHT CET 2026 · 14 MAY · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating voltage is E = 100 sin ( ω t + π 6 ) E = 100 \sin\left(\omega t + \frac{\pi}{6}\right) E = 100 sin ( ω t + 6 π )
( sin π 2 = 1 ) \left(\sin \frac{\pi}{2} = 1\right) ( sin 2 π = 1 )
(a) T 12 \frac{T}{12} 12 T (b) T 2 \frac{T}{2} 2 T (c) T 6 \frac{T}{6} 6 T (d) T 3 \frac{T}{3} 3 T Show answer and solution
Answer: (c)
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Report a mistake Q4 NEET 2026 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
The peak value of an alternating current is 5 A and frequency is 60 Hz. How long will the current, starting from zero, take to reach the peak value?
(a) 1 120 s \frac{1}{120}\text{ s} 120 1 s (b) 1 60 s \frac{1}{60}\text{ s} 60 1 s (c) 1 30 s \frac{1}{30}\text{ s} 30 1 s (d) 1 240 s \frac{1}{240}\text{ s} 240 1 s Show answer and solution
Answer: (d)
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Report a mistake Q5 MHT CET 2026 · 14 MAY · SHIFT 1 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
A resistance of 20 Ω 20\ \Omega 20 Ω V = 220 sin ( 100 π t ) V = 220 \sin(100\pi t) V = 220 sin ( 100 π t ) ( sin π 4 = 1 2 ) \left(\sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}\right) ( sin 4 π = 2 1 )
(a) 0.2 0.2 0.2 (b) 0.25 0.25 0.25 (c) 25 × 1 0 − 3 25 \times 10^{-3} 25 × 1 0 − 3 (d) 2.5 × 1 0 − 3 2.5 \times 10^{-3} 2.5 × 1 0 − 3 Show answer and solution
Answer: (d)
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Report a mistake Q6 MHT CET 2026 · 16 APR · SHIFT 1 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating e.m.f. is given by e = e 0 sin ω t e = e_0 \sin \omega t e = e 0 sin ω t
(a) T 4 \frac{T}{4} 4 T (b) T 8 \frac{T}{8} 8 T (c) T 12 \frac{T}{12} 12 T (d) T 16 \frac{T}{16} 16 T Show answer and solution
Answer: (c)
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Report a mistake Q7 MHT CET 2026 · 13 APR · SHIFT 1 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating e.m.f. is given by e = e 0 sin ω t e = e_0 \sin \omega t e = e 0 sin ω t sin 3 0 ∘ = 0.5 \sin 30^\circ = 0.5 sin 3 0 ∘ = 0.5
(a) T 4 \frac{T}{4} 4 T (b) T 8 \frac{T}{8} 8 T (c) T 12 \frac{T}{12} 12 T (d) T 16 \frac{T}{16} 16 T Show answer and solution
Answer: (c)
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Report a mistake Q8 JEE Main 2026 · 28 JAN · SHIFT 1 Difficult ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
The electric current in the circuit is given as i = i o ( t / T ) i = i_o (t/T) i = i o ( t / T ) t = 0 t = 0 t = 0 t = T t = T t = T ______ .
(a) i o 2 \frac{i_o}{\sqrt{2}} 2 i o (b) i o 3 \frac{i_o}{\sqrt{3}} 3 i o (c) i o i_o i o (d) i o 6 \frac{i_o}{\sqrt{6}} 6 i o Show answer and solution
Answer: (b)
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Report a mistake Q9 MHT CET 2025 · 19 APR · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
The instantaneous value of current in an a.c. circuit is I = 3 sin ( 50 π t + π 4 ) A I = 3 \sin\left(50\pi t + \frac{\pi}{4}\right)\text{ A} I = 3 sin ( 50 π t + 4 π ) A
(a) 1 50 s \frac{1}{50}\text{ s} 50 1 s (b) 1 100 s \frac{1}{100}\text{ s} 100 1 s (c) 1 200 s \frac{1}{200}\text{ s} 200 1 s (d) 1 600 s \frac{1}{600}\text{ s} 600 1 s Show answer and solution
Answer: (c)
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Report a mistake Q10 JEE Main 2025 · 7 APR · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
An ac current is represented as
i = 5 2 + 10 cos ( 650 π t + π 6 ) Amp i = 5\sqrt{2} + 10 \cos\left(650\pi t + \frac{\pi}{6}\right)\text{ Amp} i = 5 2 + 10 cos ( 650 π t + 6 π ) Amp
The r.m.s value of the current is
(a) 10 Amp (b) 5 2 Amp 5\sqrt{2}\text{ Amp} 5 2 Amp (c) 100 Amp (d) 50 Amp Show answer and solution
Answer: (a)
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Report a mistake Q11 JEE Main 2025 · 4 APR · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
An alternating current is represented by the equation, i = 100 2 sin ( 100 π t ) i = 100\sqrt{2} \sin(100\pi t) i = 100 2 sin ( 100 π t )
(a) 100 2 A , 100 Hz \frac{100}{\sqrt{2}}\text{ A}, 100\text{ Hz} 2 100 A , 100 Hz (b) 50 2 A , 50 Hz 50\sqrt{2}\text{ A}, 50\text{ Hz} 50 2 A , 50 Hz (c) 100 2 A , 100 Hz 100\sqrt{2}\text{ A}, 100\text{ Hz} 100 2 A , 100 Hz (d) 100 A , 50 Hz 100\text{ A}, 50\text{ Hz} 100 A , 50 Hz Show answer and solution
Answer: (d)
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Report a mistake Q12 JEE Main 2025 · 3 APR · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
An electric bulb rated as 100 W − 220 V 100\text{ W} - 220\text{ V} 100 W − 220 V
(a) 0.32 A (b) 0.64 A (c) 0.45 A (d) 2.2 A Show answer and solution
Answer: (b)
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Report a mistake Q13 JEE Main 2025 · 24 JAN · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
An alternating current is given by I = I A sin ω t + I B cos ω t I = I_A \sin \omega t + I_B \cos \omega t I = I A sin ω t + I B cos ω t
(a) I A 2 + I B 2 2 \frac{\sqrt{I_A^2 + I_B^2}}{2} 2 I A 2 + I B 2 (b) I A 2 + I B 2 2 \sqrt{\frac{I_A^2 + I_B^2}{2}} 2 I A 2 + I B 2 (c) ∣ I A + I B ∣ 2 \frac{|I_A + I_B|}{\sqrt{2}} 2 ∣ I A + I B ∣ (d) I A 2 + I B 2 \sqrt{I_A^2 + I_B^2} I A 2 + I B 2 Show answer and solution
Answer: (b)
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Report a mistake Q14 MHT CET 2025 · 22 APR · SHIFT 2 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating e.m.f. is given by e = e 0 sin ω t e = e_0 \sin \omega t e = e 0 sin ω t
( T = (T = ( T = sin 3 0 ∘ = 1 2 ) \sin 30^\circ = \frac{1}{2}) sin 3 0 ∘ = 2 1 )
(a) T 8 \frac{T}{8} 8 T (b) T 4 \frac{T}{4} 4 T (c) T 12 \frac{T}{12} 12 T (d) T 16 \frac{T}{16} 16 T Show answer and solution
Answer: (c)
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Report a mistake Q15 MHT CET 2024 · 16 MAY · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
The alternating voltage is given by v = v 0 sin ( ω t + π 3 ) v = v_0 \sin\left(\omega t + \frac{\pi}{3}\right) v = v 0 sin ( ω t + 3 π )
(a) T 6 \frac{T}{6} 6 T (b) T 3 \frac{T}{3} 3 T (c) T 2 \frac{T}{2} 2 T (d) T 12 \frac{T}{12} 12 T Show answer and solution
Answer: (d)
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Report a mistake Q16 MHT CET 2024 · 4 MAY · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating current is given by I = 100 sin ( 50 π t ) I = 100 \sin(50\pi t) I = 100 sin ( 50 π t )
(a) 50 times (b) 25 times (c) 40 times (d) 100 times Show answer and solution
Answer: (a)
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Report a mistake Q17 MHT CET 2024 · 2 MAY · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
In an a.c. circuit I = 100 sin 200 π t I = 100 \sin 200\pi t I = 100 sin 200 π t
(a) 1 100 s \frac{1}{100}\text{ s} 100 1 s (b) 1 200 s \frac{1}{200}\text{ s} 200 1 s (c) 1 300 s \frac{1}{300}\text{ s} 300 1 s (d) 1 400 s \frac{1}{400}\text{ s} 400 1 s Show answer and solution
Answer: (d)
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Report a mistake Q18 JEE Main 2024 · 4 APR · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
An alternating current at any instant is given by i = [ 6 + 56 sin ( 100 π t + π / 3 ) ] i = [6 + \sqrt{56} \sin(100\pi t + \pi/3)] i = [ 6 + 56 sin ( 100 π t + π /3 )]
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 8
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Report a mistake Q19 JEE Main 2024 · 30 JAN · SHIFT 2 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating voltage V ( t ) = 220 sin 100 π t V(t) = 220 \sin 100\pi t V ( t ) = 220 sin 100 π t 50 Ω 50\ \Omega 50 Ω
(a) 7.2 ms (b) 3.3 ms (c) 5 ms (d) 2.2 ms Show answer and solution
Answer: (b)
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Report a mistake Q20 MHT CET 2023 · 13 MAY · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
A group of lamps having total power rating of 1000 W 1000\text{ W} 1000 W E = 200 sin ( 310 t + 6 0 ∘ ) E = 200 \sin(310t + 60^\circ) E = 200 sin ( 310 t + 6 0 ∘ )
(a) 10 A 10\text{ A} 10 A (b) 5 2 A 5\sqrt{2}\text{ A} 5 2 A (c) 20 A 20\text{ A} 20 A (d) 10 2 A 10\sqrt{2}\text{ A} 10 2 A Show answer and solution
Answer: (b)
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Report a mistake Q21 MHT CET 2023 · 11 MAY · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
The a.c. source of e.m.f. with instantaneous value 'e e e e = 200 sin ( 50 t ) e = 200 \sin(50t) e = 200 sin ( 50 t ) 50 Ω 50\ \Omega 50 Ω
(a) 0.2828 A (b) 2.828 A (c) 28.28 A (d) 282.8 A Show answer and solution
Answer: (b)
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Report a mistake Q22 JEE Main 2022 · 27 JUN · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
The current flowing through an ac circuit is given by
I = 5 sin(120π \pi π
How long will the current take to reach the peak value starting from zero?
(a) 1 60 \frac{1}{60} 60 1 (b) 60 s (c) 1 120 \frac{1}{120} 120 1 (d) 1 240 \frac{1}{240} 240 1 Show answer and solution
Answer: (d)
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Report a mistake Q23 NEET 2022 · PHASE 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
A standard filament lamp consumes 100 W when connected to 200 V ac mains supply. The peak current through the bulb will be:
(a) 2 A (b) 0.707 A (c) 1 A (d) 1.414 A Show answer and solution
Answer: (b)
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Report a mistake Q24 NEET 2022 · PHASE 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
The peak voltage of the ac source is equal to
(a) The value of voltage supplied to the circuit (b) The rms value of the ac source (c) 2 \sqrt{2} 2 (d) 1/ 2 \sqrt{2} 2 Show answer and solution
Answer: (c)
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Report a mistake Q25 JEE Main 2022 · 27 JUL · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
A direct current of 4 A 4\text{ A} 4 A 4 A 4\text{ A} 4 A 3 Ω 3\ \Omega 3 Ω 2 Ω 2\ \Omega 2 Ω
(a) 3: 2 (b) 3: 1 (c) 3: 4 (d) 4: 3 Show answer and solution
Answer: (b)
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Report a mistake Q26 MHT CET 2022 · 11 AUG · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · AC meters, waveforms and skin effect
In a full wave rectifier circuit without filter, the output current is
(a) an eddy current (b) a constant direct current (c) a sinusoidal current (d) unidirectional but not steady current Show answer and solution
Answer: (d)
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Report a mistake Q27 JEE Main 2022 · 24 JUN · SHIFT 1 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
A resistance of 40 Ω \Omega Ω
(a) 2.5 ms (b) 1.25 ms (c) 2.5 s (d) 0.25 s Show answer and solution
Answer: (a)
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Report a mistake Q28 JEE Main 2022 · 27 JUN · SHIFT 1 Difficult ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
A 220 V, 50 Hz AC source is connected to a 25 V, 5 W lamp and an additional resistance R in series (as shown in figure) to run the lamp at its peak brightness, then the value of R (in ohm) will be _____________.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 975
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Report a mistake Q29 MHT CET 2021 · 24 SEP · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
The peak value of an alternating emf 'e e e e = e 0 cos ω t e = e_0 \cos \omega t e = e 0 cos ω t 50 Hz 50\text{ Hz} 50 Hz t = 1 600 s t = \frac{1}{600}\text{ s} t = 600 1 s ( cos π 6 = sin π 3 = 3 2 ) \left(\cos \frac{\pi}{6} = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\right) ( cos 6 π = sin 3 π = 2 3 )
(a) 10 V 10\text{ V} 10 V (b) 1 3 V \frac{1}{\sqrt{3}}\text{ V} 3 1 V (c) 5 V 5\text{ V} 5 V (d) 5 3 V 5\sqrt{3}\text{ V} 5 3 V Show answer and solution
Answer: (d)
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Report a mistake Q30 MHT CET 2021 · 21 SEP · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating voltge is represented by V = 80 sin ( 100 π t ) cos ( 100 π t ) V = 80 \sin(100\pi t) \cos(100\pi t) V = 80 sin ( 100 π t ) cos ( 100 π t )
(a) 20 V 20\text{ V} 20 V (b) 40 V 40\text{ V} 40 V (c) 30 V 30\text{ V} 30 V (d) 50 V 50\text{ V} 50 V Show answer and solution
Answer: (b)
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Report a mistake Q31 JEE Main 2021 · 18 MAR · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is:
(a) 2.5 ms (b) 25 ms (c) 2.5 s (d) 0.25 ms Show answer and solution
Answer: (a)
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Report a mistake Q32 MHT CET 2021 · 24 SEP · SHIFT 2 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating e.m.f. is e = e 0 sin ω t e = e_0 \sin \omega t e = e 0 sin ω t e e e T = T = T = sin 3 0 ∘ = 0.5 \sin 30^\circ = 0.5 sin 3 0 ∘ = 0.5
(a) T 12 \frac{T}{12} 12 T (b) T 16 \frac{T}{16} 16 T (c) T 4 \frac{T}{4} 4 T (d) T 8 \frac{T}{8} 8 T Show answer and solution
Answer: (a)
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Report a mistake Q33 MHT CET 2021 · 21 SEP · SHIFT 2 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
The instantaneous value of an alternating current is given by i = 50 sin ( 100 π t ) i = 50 \sin(100\pi t) i = 50 sin ( 100 π t ) 25 A 25\text{ A} 25 A ( sin 3 0 ∘ = 0.5 ) (\sin 30^\circ = 0.5) ( sin 3 0 ∘ = 0.5 )
(a) 1 300 s \frac{1}{300}\text{ s} 300 1 s (b) 1 100 s \frac{1}{100}\text{ s} 100 1 s (c) 1 200 s \frac{1}{200}\text{ s} 200 1 s (d) 1 600 s \frac{1}{600}\text{ s} 600 1 s Show answer and solution
Answer: (d)
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Report a mistake Q34 JEE Main 2021 · 25 JUL · SHIFT 2 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
A 10 Ω \Omega Ω − - −
(a) 2.5 ms (b) 1.5 ms (c) 3.0 ms (d) 4.5 ms Show answer and solution
Answer: (a)
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Report a mistake Q35 JEE Main 2021 · 17 MAR · SHIFT 1 Medium ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
An AC current is given by I = I1 sinω \omega ω ω \omega ω
(a) I 1 + I 2 2 \frac{I_1 + I_2}{\sqrt{2}} 2 I 1 + I 2 (b) I 1 2 − I 2 2 2 \sqrt{\frac{I_1^2 - I_2^2}{2}} 2 I 1 2 − I 2 2 (c) I 1 2 + I 2 2 2 \sqrt{\frac{I_1^2 + I_2^2}{2}} 2 I 1 2 + I 2 2 (d) I 1 + I 2 2 2 \frac{I_1 + I_2}{2\sqrt{2}} 2 2 I 1 + I 2 Show answer and solution
Answer: (c)
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Report a mistake Q36 JEE Main 2021 · 26 FEB · SHIFT 1 Medium ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
An alternating current is given by the equation i = i1 sin ω \omega ω ω \omega ω
(a) 1 2 ( i 1 2 + i 2 2 ) 1 2 \frac{1}{\sqrt{2}}\left(i_1^2 + i_2^2\right)^{\frac{1}{2}} 2 1 ( i 1 2 + i 2 2 ) 2 1 (b) 1 2 ( i 1 + i 2 ) \frac{1}{\sqrt{2}}(i_1 + i_2) 2 1 ( i 1 + i 2 ) (c) 1 2 ( i 1 + i 2 ) 2 \frac{1}{\sqrt{2}}(i_1 + i_2)^2 2 1 ( i 1 + i 2 ) 2 (d) 1 2 ( i 1 2 + i 2 2 ) 1 2 \frac{1}{2}\left(i_1^2 + i_2^2\right)^{\frac{1}{2}} 2 1 ( i 1 2 + i 2 2 ) 2 1 Show answer and solution
Answer: (a)
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Report a mistake Q37 JEE Main 2021 · 27 AUG · SHIFT 1 Difficult ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
The alternating current is given by i = { 42 sin ( 2 π T t ) + 10 } A i = \left\{\sqrt{42} \sin\left(\frac{2\pi}{T} t\right) + 10\right\} A i = { 42 sin ( T 2 π t ) + 10 } A
The r.m.s. value of of this current is ................. A.
Numerical answer type. Work it out, then open the answer.
Show answer and solution
Answer: 11
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Report a mistake Q38 MHT CET 2019 · 2 MAY · SHIFT 2 Easy ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating voltage is given by E = 100 sin ( ω t + π 6 ) V E = 100 \sin\left(\omega t + \frac{\pi}{6}\right)\text{ V} E = 100 sin ( ω t + 6 π ) V T = T = T =
(a) T 12 \frac{T}{12} 12 T (b) T 2 \frac{T}{2} 2 T (c) T 6 \frac{T}{6} 6 T (d) T 3 \frac{T}{3} 3 T Show answer and solution
Answer: (c)
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Report a mistake Q39 NEET 2019 Easy ☆
AC Basics: Peak, RMS & Average Values · AC meters, waveforms and skin effect
The variation of EMF with time for four types of generators are shown in the figures. Which amongst them can be called AC?
(a) Only (a) (b) (a) and (d) (c) (a), (b), (c), (d) (d) (a) and (b) Show answer and solution
Answer: (c)
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Report a mistake Q40 JEE Main 2019 · 8 APR · SHIFT 1 Medium ☆
AC Basics: Peak, RMS & Average Values · Equation, instantaneous value and time
An alternating voltage v(t) = 220 sin 100 π \pi π Ω \Omega Ω
(a) 5 ms (b) 2.2 ms (c) 3.3 ms (d) 7.2 ms Show answer and solution
Answer: (c)
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Report a mistake Q41 JEE Main 2015 · 11 APR · SHIFT 1 Easy ☆
AC Basics: Peak, RMS & Average Values · AC meters, waveforms and skin effect
The AC voltage across a resistance can be measured using a:
(a) potentiometer (b) moving coil galvanometer (c) moving magnet galvanometer (d) hot wire voltmeter Show answer and solution
Answer: (d)
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Report a mistake Q42 NEET 2011 · AIPMT · MAINS Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
The r.m.s. value of potential difference V shown in the figure is
(a) V 0 3 \frac{V_0}{\sqrt{3}} 3 V 0 (b) V0 (c) V 0 2 \frac{V_0}{\sqrt{2}} 2 V 0 (d) V 0 2 \frac{V_0}{2} 2 V 0 Show answer and solution
Answer: (c)
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Report a mistake Q43 JEE Main 2004 · AIEEE Easy ☆
AC Basics: Peak, RMS & Average Values · AC meters, waveforms and skin effect
Alternating current can not be measured by D.C. \text{D.C.} D.C.
(a) Average value of current for complete cycle is zero (b) A.C. \text{A.C.} A.C. (c) A.C. \text{A.C.} A.C. D.C. \text{D.C.} D.C. (d) D.C. \text{D.C.} D.C. Show answer and solution
Answer: (a)
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Report a mistake Q44 NEET 1994 · AIPMT Easy ☆
AC Basics: Peak, RMS & Average Values · Peak, RMS and average values
In an A.C. circuit, I rms I_{\text{rms}} I rms I 0 I_0 I 0
(a) I rms = π I 0 I_{\text{rms}} = \pi I_0 I rms = π I 0 (b) I rms = 2 I 0 I_{\text{rms}} = \sqrt{2}\, I_0 I rms = 2 I 0 (c) I rms = I 0 / π I_{\text{rms}} = I_0/\pi I rms = I 0 / π (d) I rms = I 0 / 2 I_{\text{rms}} = I_0/\sqrt{2} I rms = I 0 / 2 Show answer and solution
Answer: (d)
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Report a mistake Q45 NEET 1994 · AIPMT Easy Old syllabus ☆
AC Basics: Peak, RMS & Average Values · AC meters, waveforms and skin effect
Two cables of copper are of equal lengths. One of them has a single wire of area of cross-section A A A A / 10 A/10 A /10
(a) only multiple strands for A.C., either for D.C. (b) only multiple strands for A.C., only single strand for D.C. (c) only single strand for D.C., either for A.C. (d) only single strand for A.C., either for D.C. Show answer and solution
Answer: (a)
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