In the circuit given below,
Inductors $L_{1}$ and $L_{2}$ have resistances of $R_{1}$ and $R_{2}$ are connected with $R_{3}$ and $R_{4}$ . At t = 0, key is closed, the galvanometer always shows a zero deflection when

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$R_{1} = R_{2}$ and $\frac{R_{3}}{R_{4}} = \frac{L_{1}}{L_{2}}$

$L_{1} = L_{2}$ and $\frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$

$\frac{L_{1}}{L_{2}} = \frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$

$\frac{L_{1}}{L_{2}} = \frac{R_{2}}{R_{1}} = \frac{R_{3}}{R_{4}}$

answer is C.

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Detailed Solution

Given circuit is,

As galvanometer shows no deflection, $V_{A} = V_{B}$ .

So, $V_{P A} = V_{P B}$

$\begin{array}{rcl} L_{1} \frac{d i_{1}}{d t} + i_{1} R_{1} & = & L_{2} \frac{d i_{2}}{d t} + i_{2} R_{2} . . . . . . (i) \\ and V_{A Q} & = & V_{B Q} \\ i_{1} R_{3} & = & i_{2} R_{4} . . . . . . (i i) \end{array}$

Substituting for $i_{1}$ in equation (i) from equation (ii) we get,

$L_{1} \frac{R_{4}}{R_{3}} \frac{d i_{2}}{d t} + i_{2} \frac{R_{1}}{R_{3}} = L_{2} \frac{d i_{2}}{d t} + i_{2} R_{2}$

At $t = 0, i_{2} = 0$ .

So, $L_{1} \frac{R_{4}}{R_{3}} - L_{2} = 0 \Rightarrow \frac{L_{1}}{L_{2}} = \frac{R_{3}}{R_{4}}$

At $t = \infty$ ,

$\frac{d i_{2}}{d t} = 0 \Rightarrow i_{2} = \frac{V}{R_{2} + R_{4}} \Rightarrow R_{2} - \frac{R_{4} R_{1}}{R_{3}} = 0 \Rightarrow \frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$