A coil of self inductance $2 \cdot 5 H$ and resistance $20 Ω$ is connected to a battery of emf 120V having internal resistance of $5 Ω$ . Find the current in the circuit in steady state.

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Solution

Current in the circuit $i_{0} = \frac{E M F}{internal resistance} = \frac{E}{R} = \frac{120}{20}$
$= 6 A$

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A coil of self inductance 2⋅5H and resistance 20Ω is connected to a battery of emf 120V having internal resistance of 5Ω. Find the current in the circuit in steady state.

Current in the circuit i0=EMFinternal resistance=ER=12020=6A

A coil of self inductance $2 \cdot 5 H$ and resistance $20 Ω$ is connected to a battery of emf 120V having internal resistance of $5 Ω$ . Find the current in the circuit in steady state.

Current in the circuit $i_{0} = \frac{E M F}{internal resistance} = \frac{E}{R} = \frac{120}{20}$
$= 6 A$