Question

In the bridge circuit shown, the capacitors are loss free. At balance, the value of capacitance $C_1$ in microfarad is

Answer 1

Redrawing the given bridge circuit, we have:
$Z_1=\frac{1}{j\omega C_1};\Omega Z_2=35k\Omega$
$Z_3=\frac{{10}^6}{j0.1\omega }\Omega ,Z_4=105k\Omega$
At balance, current though galvanometer;
$I_g$=0
and $|Z_1||Z_4|=|Z_2||Z_3|$
$\therefore \left(\frac{1}{\omega C_1}\right)\times \left(105K\right)$ =$\left(35K\right)\left(\frac{{10}^6}{0.1\omega }\right)$
or, $C_1=\frac{105\times 0.1}{35}\mu F$
$=\frac{10.5}{35}\mu F=0.3\mu F$
$\therefore In\mu F,$ $C_1=0.3$