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}

Heaciside Campbell bridge method is commonly used for finding mutual inductance.

\begin{aligned} C_P &=0.102\mu F \\ R_P &= 1.25M\Omega \\ \end{aligned}
\begin{aligned} \text{Power loss}&=\frac{V^2}{R_P} =\frac{(5 \times 10^3)^2}{1.25 \times 10^6}=\frac{25}{1.25}\\ &=20W \\ \tan\delta &=RIX=\frac{1}{\omega C_PR_P} \\ &=\frac{1}{2 \times \pi \times 50 \times 0.102 \times 10^{-6}\times 1.25 \times 10^6} \\ &= 0.025 \end{aligned}

The bridge is Maxwell bridge.
Element is an inductor.
Inductance =L_x effective resistance of the inductor =R_x
Q=\frac{\omega L_x}{R_x}=\omega C_1R_1 \;\;...(i)
The bridge is limited to measurement of low Q inductor (1 \lt Q \lt 10).
It is clear from equation (i) that the measurement of high Q coils demands a large value of resistance R_1, perhaps 10^5 or 10^6 \Omega . The resistance boxes of such high value are very expensive. Thus for value of Q \gt 10, the bridge is unsuitable.
The bridge is also unsuited for coils with a very low value of Q(i.e. Q \lt 1).

Z_1=R+j\omega L
Z_2=R_2
Z_3=R_3
Z_4=R_4||\left ( \frac{-j}{\omega C_4} \right )
\;\;=R_4||\left ( \frac{1}{\omega C_4} \right )
\;\;=\frac{R_4}{1+j\omega C_4R_4}
At balance,
Z_1Z_4=Z_2Z_3
(R+ j\omega L)\left ( \frac{R_4}{1+j\omega C_4R_4} \right )=R_2R_3
R_4(R+ j\omega L)=R_2R_3(1+j\omega C_4R_4)
RR_4+j\omega L R_4=R_2R_3+j\omega R_2R_3R_4C_4
Equating real and imaginary terms
RR_4=R_2R_3
R=\frac{R_2R_3}{R_4}
\omega LR_4=\omega R_2R_3R_4C_4
L=R_2R_3C_4

Z_{AB}=500\Omega
Z_{CD}=Z
Z_{BC}=R_{BC}+\frac{1}{j\omega C}
\;\;=300-\frac{j}{2 \pi \times 2 \times 10^3 \times 0.398 \times 10^{-6}}
Z_{BC}=300-j200\Omega
Z_{AD}=R_{AD}+j\omega L
Z_{AD}=300+j2\pi\times 2 \times 10^3 \times 15.91 \times 10^{-3}
\;\;=300+j200\Omega
At balance,
Z_{AB} \times Z_{CD}=Z_{BC} \times Z_{AD}
Z_{CD}=Z
\;\;=\frac{Z_{BC} \times Z_{AD}}{Z_{AB}}
\Rightarrow \;Z=\frac{(300-j200)\times (300+j200)}{500}
Z=(260+j0)\Omega