Alternating Current | Physics JEE Main questions with Solutions

Q 1 :

An alternating current at any instant is given by $i = [6 + \sqrt{56} \sin (100 π t + \frac{π}{3})] A$ . The rms value of the current is _______ A [2024]

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(8) $If i = a + b \sin (ω t + θ)$

$i_{rms} = \sqrt{a^{2} + \frac{b^{2}}{2}}$

$∴ i_{rms} = \sqrt{36 + \frac{56}{2}}$

$= \sqrt{36 + 28} = \sqrt{64} = 8 A$

Q 2 :

An alternating voltage $V (t) = 220$ $\sin$ $100 π t$ volt is applied to a purely resistive load of 50 $Ω$ . The time taken for the current to rise from half of the peak value to the peak value is [2024]

5 ms
3.3 ms
7.2 ms
2.2 ms

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(2)

$V (t) = 220 \sin (100 π t)$

$I = \frac{V_{s}}{R} = \frac{220}{50} \sin (100 π t)$

$I = \frac{22}{5} \sin (100 π t)$

$\cos θ = \frac{\frac{I_{0}}{2}}{I_{0}} = \frac{1}{2} \Rightarrow θ = 60 ° = \frac{π}{3}$

Using phasor, $t = \frac{π / 3}{100 π}$

$t = \frac{1}{300} sec = \frac{10}{3} msec$

$t = 3.3 msec$

Q 3 :

An alternating current is given by

$I = I_{A} \sin ω t + I_{B} \cos ω t$

The r.m.s. current will be [2025]

$\sqrt{I_{A}^{2} + I_{B}^{2}}$
$\frac{\sqrt{I_{A}^{2} + I_{B}^{2}}}{2}$
$\sqrt{\frac{I_{A}^{2} + I_{B}^{2}}{2}}$
$\frac{| I_{A} + I_{B} |}{\sqrt{2}}$

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(3)

$I = I_{A} \sin ω t + I_{B} \cos ω t$

So, $I_{RMS} = \sqrt{\frac{I_{A}^{2} + I_{B}^{2}}{2}}$

Q 4 :

An electric bulb rated as 100 W – 220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is: [2025]

0.64 A
0.45 A
2.2 A
0.32 A

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(1)

$R_{B} = \frac{V^{2}}{P} = \frac{220 \times 220}{100} = 484 Ω$

$I_{rms} = \frac{V_{rms}}{R} = \frac{220}{484} = \frac{10}{22} = \frac{5}{11} A$

$I_{0} = \frac{5 \sqrt{2}}{11} = 0.64 A$

Q 5 :

An alternating current is represented by the equation, $i = 100 \sqrt{2} \sin (100 π t)$ ampere. The RMS value of current and the frequency of the given alternating current are [2025]

$100 \sqrt{2} A, 100 Hz$
$\frac{100}{\sqrt{2}} A, 100 Hz$
100 A, 50 Hz
$50 \sqrt{2} A, 50 Hz$

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(3)

$i = 100 \sqrt{2} \sin (100 π t)$

$i_{r m s} = \frac{100 \sqrt{2}}{\sqrt{2}} = 100 A$

$f = \frac{ω}{2 π} = \frac{100 π}{2 π} = 50 Hz$

Q 6 :

An ac current is represented as

$i = 5 \sqrt{2} + 10 \cos (650 π t + \frac{π}{6}) Amp$

The r.m.s. value of the current is [2025]

20 Amp
100 Amp
10 Amp
$5 \sqrt{2} Amp$

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(3)

$i = 5 \sqrt{2} + 10 \cos (650 π t + \frac{π}{6})$

$i^{2} = 50 + 100 \cos^{2} (650 π t + \frac{π}{6})$ $+ (2) (5 \sqrt{2}) (10) \cos (650 π t + \frac{π}{6})$

$< i^{2} > = 50 + \frac{100}{2} + 0$

$< i^{2} > = 100$

$< i > = 10 Amp$

Q 7 :

Match the List-I with List-II. [2023]

List I List II

A. AC generator I. Presence of both L and C

B. Transformer II. Electromagnetic Induction

C. Resonance phenomenon to occur III. Quality factor

D. Sharpness of resonance IV. Mutual Inductance

Choose the correct answer from the options given below:

A-IV, B-II, C-I, D-III
A-II, B-I, C-III, D-IV
A-II, B-IV, C-I, D-III
A-IV, B-III, C-I, D-II

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(3)

AC generator works on Electromagnetic principle (A–II)

Transformer works on Mutual induction (B–IV)

Resonance occurs when both L and C are present (C–I) and

quality factor determines sharpness of resonance (D–III)

Q 8 :

A capacitor of capacitance $150.0 μ F$ is connected to an alternating source of emf given by $E = 36 \sin (120 π t) V .$ The maximum value of current in the circuit is approximately equal to [2023]

$2 A$
$\frac{1}{\sqrt{2}} A$
$\sqrt{2} A$
$2 \sqrt{2} A$

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(1)

$Maximum current,$

$I_{0} = \frac{E_{0}}{X_{c}} = \frac{E_{0}}{\frac{1}{ω C}} = E_{0} ω C$

$\Rightarrow I_{0} = 36 \times 120 π \times 150 \times 10^{- 6} = 2 A$

Q 9 :

As per the given graph choose the correct representation for curve A and curve B. [2023]

{Where: $X_{C}$ = reactance of pure capacitive circuit connected with A.C. source

$X_{L}$ = reactance of pure inductive circuit connected with A.C. source

$R$ = impedance of pure resistive circuit connected with A.C. source

$Z$ = impedance of the LCR series circuit}

$A = X_{C}, B = R$
$A = X_{L}, B = Z$
$A = X_{C}, B = X_{L}$
$A = X_{L}, B = R$

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(3)

$X_{C} = \frac{1}{ω C} = \frac{1}{(2 π f) C}$

$∴$ $X_{C} \propto \frac{1}{f}$

$∴$ $Curve A$

$X_{L} = ω L = (2 π f) L$

$∴$ $X_{L} \propto f$

$∴$ $Curve B$

Q 10 :

The electric current in the circuit is given as $i = i_{0} ((\frac{t}{T})) .$ The r.m.s current for the period $t = 0 t o t = T$ is_______. [2026]

$\frac{i_{0}}{\sqrt{3}}$
$\frac{i_{0}}{\sqrt{2}}$
$\frac{i_{0}}{\sqrt{6}}$
$i_{0}$

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Topic Question Set

Q 1 :

An alternating current at any instant is given by $i = [6 + \sqrt{56} \sin (100 π t + \frac{π}{3})] A$ . The rms value of the current is _______ A [2024]

0%

Q 2 :

An alternating voltage $V (t) = 220$ $\sin$ $100 π t$ volt is applied to a purely resistive load of 50 $Ω$ . The time taken for the current to rise from half of the peak value to the peak value is [2024]

Q 3 :

An alternating current is given by

$I = I_{A} \sin ω t + I_{B} \cos ω t$

The r.m.s. current will be [2025]

Q 4 :

An electric bulb rated as 100 W – 220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is: [2025]

Q 5 :

An alternating current is represented by the equation, $i = 100 \sqrt{2} \sin (100 π t)$ ampere. The RMS value of current and the frequency of the given alternating current are [2025]

Q 6 :

An ac current is represented as

$i = 5 \sqrt{2} + 10 \cos (650 π t + \frac{π}{6}) Amp$

The r.m.s. value of the current is [2025]

Q 8 :

A capacitor of capacitance $150.0 μ F$ is connected to an alternating source of emf given by $E = 36 \sin (120 π t) V .$ The maximum value of current in the circuit is approximately equal to [2023]

Q 10 :

The electric current in the circuit is given as $i = i_{0} ((\frac{t}{T})) .$ The r.m.s current for the period $t = 0 t o t = T$ is_______. [2026]

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	List I		List II
A.	AC generator	I.	Presence of both L and C
B.	Transformer	II.	Electromagnetic Induction
C.	Resonance phenomenon to occur	III.	Quality factor
D.	Sharpness of resonance	IV.	Mutual Inductance