An inductor of inductance- L-5 H is connected to an AC source having voltage- V-10sin10t-6- Find the peak value of current in the circuit-

Given: Inductance, nbsp;L=5 H Voltage, nbsp;V=10sin(10t+π6) So, peak value of current in the circuit, i_0=V_0X_L=V_0ω L=1010×5=0.2 A Hence, option (B) is ...

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Standard XII

Physics

Alternating Current Applied to an Inductor

An inductor o...

Question

An inductor of inductance, $L = 5 H$ is connected to an AC source having voltage, $V = 10 sin (10 t + \frac{π}{6})$ . Find the peak value of current in the circuit.

0.3 A

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0.2 A

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0.4 A

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0.1 A

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Solution

The correct option is B $0.2 A$
Given:
Inductance, $L = 5 H$
Voltage, $V = 10 sin (10 t + \frac{π}{6})$

So, peak value of current in the circuit,

$i_{0} = \frac{V_{0}}{X_{L}} = \frac{V_{0}}{ω L} = \frac{10}{10 \times 5} = 0.2 A$

Hence, option $(B)$ is the correct answer.

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An inductor of inductance, L=5 H is connected to an AC source having voltage, V=10sin(10t+π6). Find the peak value of current in the circuit.

The correct option is B 0.2 AGiven: Inductance, L=5 H Voltage, V=10sin(10t+π6) So, peak value of current in the circuit, i0=V0XL=V0ωL=1010×5=0.2 A Hence, option (B) is the correct answer.

An inductor of inductance, $L = 5 H$ is connected to an AC source having voltage, $V = 10 sin (10 t + \frac{π}{6})$ . Find the peak value of current in the circuit.

The correct option is B $0.2 A$
Given:
Inductance, $L = 5 H$
Voltage, $V = 10 sin (10 t + \frac{π}{6})$

So, peak value of current in the circuit,

$i_{0} = \frac{V_{0}}{X_{L}} = \frac{V_{0}}{ω L} = \frac{10}{10 \times 5} = 0.2 A$

Hence, option $(B)$ is the correct answer.