Question

Consider the AC bridge shown below. If $\omega RC=1and\Delta C/C<0.01,$ then the ratio $\left|\frac{V_0}{V_s}\right|$ is approximately equal to

• A. $\sqrt{1+\frac{\Delta C}{C}}$
• B. $\frac{2\Delta C}{C}$
• C. $\frac{\Delta C}{C}$
• D. $\frac{\Delta C}{2C}$

Answer 1

The correct option is D $\frac{\Delta C}{2C}$


$E_{ab}=\frac{R}{R+\frac{1}{\omega (C+\Delta C)}}{V}_s$

$=\frac{R\omega (C+\Delta C)}{R\omega (C+\Delta C)+1}{V}_s$

$E_{ad}=\frac{R}{R+\frac{1}{\omega C}}{V}_s=\frac{R\omega C}{R\omega C+1}{V}_s$

$V_0=E_{ab}-E_{ad}$

$=\frac{R\omega (C+\Delta C)}{R\omega (C+\Delta C)+1}{V}_s-\frac{R\omega C}{R\omega C+1}{V}_s$

$V_0=[\frac{R\omega C+R\omega \Delta C}{R\omega C+R\omega \Delta C+1}-\frac{R\omega C}{R\omega C+1}]V_s$

$\frac{V_o}{V_s}=\frac{1+R\omega \Delta C}{1+1+R\omega \Delta C}-\frac{1}{1+1}$

$=\frac{1+R\omega \Delta C}{2+R\omega \Delta C}-\frac{1}{2}$

$=\frac{1+1+R\omega \Delta C-1}{2+R\omega \Delta C}-\frac{1}{2}$

$=1-\frac{1}{2+R\omega \Delta C}-\frac{1}{2}$

$=\frac{1}{2}-\frac{1}{2+R\omega C\times \left(\frac{\Delta C}{C}\right)}$

$=\frac{1}{2}-\frac{1}{2+\frac{\Delta C}{C}}=\frac{2+\frac{\Delta C}{C}-2}{2+\frac{\Delta C}{C}}$

$\frac{\Delta C}{C}<<0.01,$

$\therefore$ it can be neglected in comparison to 2

$=\frac{\Delta C}{\frac{C}{2}}=\frac{\Delta C}{2C}$