Alternating Current | Physics JEE Main questions with Solutions

Q 1 :
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb [2024]

increases
remains same
becomes zero
decreases

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4

(A) As dielectric is placed, capacitance will increase.

$X_{C} = \frac{1}{C ω}$

It means the capacitive reactance decreases,so impedance of circuit decreases. Hence current increases. So power increases. So glow of the bulb increases.

Q 2 :
In an a.c. circuit, voltage and current are given by:

$V = 100 \sin (100 t) V$

and $I = 100 \sin (100 t + \frac{π}{3}) m A$ respectively.

The average power dissipated in one cycle is [2024]

10 W
2.5 W
25 W
5 W

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(B) $P_{avg} = V_{rms}$ $I_{rms} \cos (Δ ϕ)$

$= \frac{100}{\sqrt{2}} \times \frac{100 \times 10^{- 3}}{\sqrt{2}} \times \cos (\frac{π}{3}) = 2.5 W$

Q 3 :
A series L, R circuit connected with an ac source $E = (25$ $\sin$ $1000 t)$ $V$ has a power factor of $\frac{1}{\sqrt{2}}$ . If the source of emf is changed to $E = (20$ $\sin$ $2000 t)$ $V$ , the new power factor of the circuit will be [2024]

$\frac{1}{\sqrt{2}}$
$\frac{1}{\sqrt{3}}$
$\frac{1}{\sqrt{5}}$
$\frac{1}{\sqrt{7}}$

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(C) For first source power factor

$\cos ϕ = \frac{R}{Z} = \frac{R}{\sqrt{R^{2} + ω^{2} L^{2}}}$

$\Rightarrow \frac{1}{\sqrt{2}} = \frac{R}{\sqrt{R^{2} + ω^{2} L^{2}}} \Rightarrow R^{2} + ω^{2} L^{2} = 2 R^{2}$

$\Rightarrow R = ω L$

for second source, $ω^{'} = 2000 = 2 ω$

Power Factor = $\frac{R}{\sqrt{R^{2} + ω^{' 2} L^{2}}}$

$= \frac{ω L}{\sqrt{ω^{2} L^{2} + 4 ω^{2} L^{2}}} = \frac{ω L}{\sqrt{5 ω^{2} L^{2}}} = \frac{1}{\sqrt{5}}$

Q 4 :
An AC voltage V = 20 sin 200 $π t$ is applied to a series LCR circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is [2024]

21.6 W
200 W
173.2 W
50 W

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(D) $< P > = I_{rms}$ $V_{rms} \cos ϕ$

$= \frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60 ° = 50 W$

Topic Question Set

Q 1 :
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb [2024]

Q 2 :
In an a.c. circuit, voltage and current are given by:

$V = 100 \sin (100 t) V$

and $I = 100 \sin (100 t + \frac{π}{3}) m A$ respectively.

The average power dissipated in one cycle is [2024]

Q 3 :
A series L, R circuit connected with an ac source $E = (25$ $\sin$ $1000 t)$ $V$ has a power factor of $\frac{1}{\sqrt{2}}$ . If the source of emf is changed to $E = (20$ $\sin$ $2000 t)$ $V$ , the new power factor of the circuit will be [2024]

Q 4 :
An AC voltage V = 20 sin 200 $π t$ is applied to a series LCR circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is [2024]

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Topic Question Set

Q 1 : A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb [2024]

Q 2 : In an a.c. circuit, voltage and current are given by: V=100sin(100t) V and I=100sin(100t+π3)mA respectively. The average power dissipated in one cycle is [2024]

Q 3 : A series L, R circuit connected with an ac source E=(25 sin 1000t)V has a power factor of 12. If the source of emf is changed to E=(20 sin 2000t)V, the new power factor of the circuit will be [2024]

Q 4 : An AC voltage V = 20 sin 200πt is applied to a series LCR circuit which drives a current I=10sin(200πt+π3). The average power dissipated is [2024]

Want Us to Email you About Special Offers & Updates?

Q 1 :
A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb [2024]

Q 2 :
In an a.c. circuit, voltage and current are given by:

$V = 100 \sin (100 t) V$

and $I = 100 \sin (100 t + \frac{π}{3}) m A$ respectively.

The average power dissipated in one cycle is [2024]

Q 3 :
A series L, R circuit connected with an ac source $E = (25$ $\sin$ $1000 t)$ $V$ has a power factor of $\frac{1}{\sqrt{2}}$ . If the source of emf is changed to $E = (20$ $\sin$ $2000 t)$ $V$ , the new power factor of the circuit will be [2024]

Q 4 :
An AC voltage V = 20 sin 200 $π t$ is applied to a series LCR circuit which drives a current $I = 10 \sin (200 π t + \frac{π}{3})$ . The average power dissipated is [2024]