Question

A capacitor of capacitance $C$ is charged to voltage $V_0$ and allowed to discharge through a resistance R while charge another capacitance${}^{\prime }\alpha C^{\prime }$. What fraction of total energy is lost at steady state?

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

Answer 1

The correct option is B $\frac{1}{1+\frac{1}{\alpha }}$



Let $V_f$ be the final voltage appearing across $ab$
Equating charge of two capacitors,
$V_f\alpha C=V_{0}C-V_{f}C$
$V_f(1+\alpha )=V_0$
$V_f=\frac{V_0}{1+\alpha }$
Initial energy $=\frac{1}{2}C{V}_0^2$
Final energy $=\frac{1}{2}\left(\alpha C\right)V_f^2(1+\alpha )=\frac{1}{2}C(1+\alpha ){\left(\frac{V_0}{1+\alpha }\right)}^2$
$=\frac{1}{2}C{V}_0^2\frac{1}{1+\alpha }$
Loss in energy = inital energy - final energy
$=\frac{1}{2}C{V}_0^2-\frac{1}{2}C{V}_0^2\left(\frac{1}{1+\alpha }\right)$
loss in energy $=\frac{1}{2}C{V}_0^2\left(\frac{\alpha }{1+\alpha }\right)$
The fraction of energy lost$=\left(\frac{\alpha }{1+\alpha }\right)=\frac{1}{1+\frac{1}{\alpha }}$