In the figure shown, a circuit contains two identical resistors with resistance $R = 5Ω$ and an inductance with $L = 2 m H .$ An ideal battery of $15 V$ is connected in the circuit. What will be the current through the battery long after the switch is closed?

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TestHub · Accepted Answer

After a long time, when the growth of current in the circuit has stopped, the inductor will work as a conductor. Therefore the effective resistance of the circuit will be,

$R = \frac{5}{2} = 2.5Ω$

Hence, the current through the circuit will be,

$I = \frac{V}{R} = \frac{15}{2.5} = 6 A$

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