Question

The inductance of a coil is measured using the bridge shown in the figure. Balance (D = 0) is obtained with $C_1=1nF,R_1=2.2M\Omega ,R_2=22.2k\Omega ,R_4=10k\Omega .$ The value of the inductance $L_x$ (in mH) is

• A. 222

Answer 1

The correct option is A 222
In the given bridge circuit,

$Z_1=\frac{R_1}{1+j\omega C_1{R}_1}$

$Z_2=R_2$

$Z_3=R_x+j\omega L_x$

$Z_4=R_4$

(2) balance product of impedance of opposite arms are equal.

i.e. $Z_1{Z}_3=Z_2{Z}_4$

$\frac{R}{1+j\omega C_1{R}_1}(R_x+j\omega L_x)=R_2{R}_4$

$R_1{R}_x+j\omega R_1{L}_x=R_2{R}_4+j\omega C_1{R}_1{R}_2{R}_4$

Equating the imaginary parts

$L_x=C_1{R}_2{R}_4$

Substituting the values we have

$L_x=1\times {10}^{-9}\times 22.2\times {10}^3\times 10\times {10}^3$

$=222mH$