Two inductors of self-inductances $L_{1}$ and $L_{2}$ and of resistances $R_{1}$ and $R_{2}$ (not shown here) respectively connected in the circuit as shown in Fig. At the instant $t = 0,$ key $K$ is closed. Obtain an expression which the galvanometer will show zero deflection at all times after the key is closed.

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

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

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

None of these

answer is A.

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Detailed Solution

Since there is no current through $B D$ therefore points $B$ and $D$ are at the same potential.
Potential difference, $V_{B} = V_{D}$
$V_{A B} = V_{A D}$

Two inductors of self inductances L_(1) and L_(2) of reistances R_(1) and R _(2) (not shown in the diagram) respectively, are connected in the circuit as shown in the figure. At the instant

$L_{1} \frac{d i_{1}}{d t} + i_{1} R_{1} = L_{2} \frac{d i_{2}}{d t} + i_{2} R_{2} \dots . (1)$
Similarly, $V_{B C} = V_{D C} \dots \dots \dots \dots \dots \dots . . (2)$
From equations $(1)$ and $(2)$ ,
$\frac{d i_{1}}{d t} R_{3} = \frac{d i_{2}}{d t} R_{4}$
$\frac{d i_{1}}{d t} = \frac{R_{4}}{R_{3}} \frac{d i_{2}}{d t}$
$i_{1} = i_{2} \frac{R_{4}}{R_{3}}$
$L_{1} \frac{R_{4}}{R_{3}} \frac{d i_{2}}{d t} + i_{2} \frac{R_{1} R_{4}}{R_{3}} = L_{2} \frac{d i_{2}}{d t} + i_{2} R_{2}$
$(L_{1} \frac{R_{4}}{R_{3}} - L_{2}) \frac{d i_{2}}{d t} = i_{2} [R_{2} - \frac{R_{1} R_{4}}{R_{3}}] \dots \dots \dots \dots . . (3)$
At $t = 0, i_{2} = 0$
$∴ L_{1} \frac{R_{4}}{R_{3}} - L_{2} = 0 as \frac{d i_{2}}{d t} \neq 0$
$∴ \frac{L_{1}}{L_{2}} = \frac{R_{3}}{R_{4}} \dots \dots \dots \dots \dots . (4)$
At $t = \infty, \frac{d i_{2}}{d t} = 0, i_{2} = (\frac{E}{R_{2} + R_{4}}) = constant$
$\Rightarrow R_{2} - \frac{R_{4} R_{1}}{R_{3}} = 0$
$∴ \frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}} \dots \dots \dots \dots \dots (5)$
From equations $(4) and (5), \frac{L_{1}}{L_{2}} = \frac{R_{1}}{R_{2}} = \frac{R_{3}}{R_{4}}$