In the circuit shown below, i 1 and i 2 are the steady-state values of the current through L 1 and L 2 respectively, then i 1 is

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TestHub · Accepted Answer

Since the potential difference across the inductors is same, we get

$L_{1} \frac{d i_{1}}{d t} = L_{2} \frac{d i_{2}}{d t}$

$L_{1} d i_{1} = L_{2} d i_{2}$

$L_{1} ∆ i_{1} = L_{2} ∆ i_{2}$

$L_{1} i_{1} = L_{2} i_{2}$ (Since the initial value of current through both the inductors was zero)

$\frac{i_{1}}{i_{2}} = \frac{L_{2}}{L_{1}}$

The steady current $i$ passing through $R$ is

$i = \frac{E}{R}$

$i_{1} = (\frac{L_{2}}{L_{1} + L_{2}}) i = \frac{E L_{2}}{R (L_{1} + L_{2})}$

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