A.C. Bridges

The condition for balanced bridge
\begin{aligned} R_{AB}R_{CD}&=R_{DA}R_{BC} \\ R_{CD} &=\frac{300 \times 100}{1000}=30\Omega \\ %Error &=\pm (2.1+0.5+0.4)=\pm 3% \\ \therefore \; R_{CD}&=30\pm 30 \times \frac{3}{100}=30\pm 0.9\Omega \end{aligned}

Maxwell's bridge is used for measurement of inductance.
Wein's bridge is used for measurement of frequency
Kelvin's bridge is used for measurement of low value of resistance.
Schering bridge is used for measurement of capacitance, dilectric loss and permittivity etc.

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

Given bridge is shown in figure

From the given bridge circuit, we see that product of opposite arm impedance are equal
\begin{aligned} i.e. \;\;& (-j) \times (-j)=(j)(j)\\ or\;\;\; & -1= -1 \\ \end{aligned}
Hence, the bridge is balanced, i.e. no current flows through the voltage source V.
Now, \begin{aligned} v_i(t)&= 100 \sin \omega t\\ &= \frac{100}{\sqrt{2}} \text{Volt(rms value)}\\ \end{aligned}
\therefore Net current supplied by the source is
I=\frac{100\sqrt{2}}{(0.5+Z_{eq})}A
Z_{eq}=0 (i.e. net impedance of whole bridge=0)
\therefore \;\; I=\frac{100\sqrt{2}}{0.5}=141.42A
\therefore \;\;\frac{I_2}{2}=70.71A
Current flow through each parallel path of the bridge circuit, therefore V=V_a-V_b
\;\;=\left ( \frac{I}{2} \times -j \right )-\left ( \frac{I}{2} \times j \right )
\;\;=-jI=-j141.42 volt (rms value)
\therefore Reading of the voltmeter in volt (rms)
=V=V_{ab}=141.42

\begin{aligned} V_1&= -|j1\Omega | \times I \\ V_2 &= |j2\Omega | \times I \\ V&= -j1\Omega | \times I +j2\Omega | \times I \\ &=V_2-V_1 \end{aligned}