Question

The parallel resistance-capacitance bridge shown below has a standard capacitance value of $C_1=0.1\mu F$ and a resistance value of $R_3=10k\Omega .$ The bridge is balanced at a supply frequency of 100 Hz for$R_1=375k\Omega ,R_3=10k\Omega \text{and}{R}_4=14.7k\Omega .$ The value of the dissipation factor $D=\frac{1}{\left(\omega R_p{C}_p\right)}$ of the parallel combination of Cp and Rp is ____ . (Answer should be rounded off to Three decimal places).

• A. 0.042

Answer 1

The correct option is A 0.042
In the given bridge circuit,

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

$Z_2=\frac{R_P}{1+j\omega C_P{R}_P}$

$Z_3=R_3$

$Z_4=R_4$

At balance condition,

$Z_1{Z}_4=Z_2{Z}_3$

$\frac{R_1{R}_4}{1+j\omega C_1{R}_1}=\frac{R_P{R}_3}{1+j\omega C_P{R}_P}$

$R_1{R}_4+j\omega C_P{R}_P{R}_1{R}_4=R_P{R}_3+j\omega C_1{R}_1{R}_P{R}_3$

Equating the real and imaginary components

$R_1{R}_4=R_P{R}_{3}or{R}_P=\frac{R_1{R}_4}{R_3}$

$C_P{R}_4=C_1{R}_3$

$or{C}_P=\frac{C_1{R}_3}{R_4}$

Given,

$D=\frac{1}{\omega C_P{R}_P}=\frac{1}{\omega }\times \frac{R_4}{C_1{R}_3}\times \frac{R_3}{R_1{R}_4}=\frac{1}{\omega C_1{R}_1}$

Given,

$C_1=0.1\times {10}^{-6}F$

$R_1=375\times {10}^3\Omega$

$\omega =2\times \pi \times 100=628.318$

$\therefore D=\frac{1}{628.318\times 0.1\times {10}^{-6}\times 375\times {10}^3}\simeq 0.042$