In the circuit shown below, the inductance is connected to an AC source. The current flowing in the circuit is . The velocity drop across is [2024]

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Q.
In the circuit shown below, the inductance $L$ is connected to an AC source. The current flowing in the circuit is $I = I_{0} \sin ω t$ . The velocity drop $(V_{L})$ across $L$ is [2024]

1 $ω L I_{0} \sin ω t$

2 $\frac{I_{0}}{ω L} \sin ω t$

3 $\frac{I_{0}}{ω L} \cos ω t$

4 $ω L I_{0} \cos ω t$

Ans.
(4)

Given, $I = I_{0} \sin ω t$

The voltage drop $(V_{L})$ across inductor is given by,

$V_{L} = L \frac{d I}{d t} = L \frac{d}{d t} (I_{0} \sin ω t)$

$V_{L} = ω L I_{0} \cos ω t$

Solution

Q.
In the circuit shown below, the inductance $L$ is connected to an AC source. The current flowing in the circuit is $I = I_{0} \sin ω t$ . The velocity drop $(V_{L})$ across $L$ is [2024]

1 $ω L I_{0} \sin ω t$

2 $\frac{I_{0}}{ω L} \sin ω t$

3 $\frac{I_{0}}{ω L} \cos ω t$

4 $ω L I_{0} \cos ω t$

Ans.
(4)

Given, $I = I_{0} \sin ω t$

The voltage drop $(V_{L})$ across inductor is given by,

$V_{L} = L \frac{d I}{d t} = L \frac{d}{d t} (I_{0} \sin ω t)$

$V_{L} = ω L I_{0} \cos ω t$

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Solution

Q. In the circuit shown below, the inductance L is connected to an AC source. The current flowing in the circuit is I=I0sinωt. The velocity drop (VL) across L is [2024]

1 ωLI0sinωt

2 I0ωLsinωt

3 I0ωLcosωt

4 ωLI0cosωt

Ans. (4) Given, I=I0sinωt The voltage drop (VL) across inductor is given by, VL=LdIdt=Lddt(I0sinωt) VL=ωLI0cosωt

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Q.
In the circuit shown below, the inductance $L$ is connected to an AC source. The current flowing in the circuit is $I = I_{0} \sin ω t$ . The velocity drop $(V_{L})$ across $L$ is [2024]

1 $ω L I_{0} \sin ω t$

2 $\frac{I_{0}}{ω L} \sin ω t$

3 $\frac{I_{0}}{ω L} \cos ω t$

4 $ω L I_{0} \cos ω t$

Ans.
(4)

Given, $I = I_{0} \sin ω t$

The voltage drop $(V_{L})$ across inductor is given by,

$V_{L} = L \frac{d I}{d t} = L \frac{d}{d t} (I_{0} \sin ω t)$

$V_{L} = ω L I_{0} \cos ω t$