Question

A capacitor $C_1$ is charged to a potential $V$ and connected to another capacitor in series with a resistor $R$ as shown. It is observed that heat $H_1$ is dissipated across resistance $R$, till the circuit reaches steady state. Same process is repeated using resistance of $2R$. If $H_2$ is the heat dissipated in this case then:

• A. $\frac{H_2}{H_1}=1$
• B. $\frac{H_2}{H_1}=4$
• C. $\frac{H_2}{H_1}=\frac{1}{4}$
• D. $\frac{H_2}{H_1}=2$

Answer 1

Answer: (A)

$\frac{H_2}{H_1}=1$

We know that heat produced$\left(H\right)$ is equal to the energy loss $(U_i-U_f)$.
Here in both case the capacitors remain unchanged, so energy loss is also unchanged.
So $H_1=H_2=U_i-U_f$
$\therefore \frac{H_2}{H_1}=1$
The only change is by increasing the resistance. This makes the time constant to increase. Hence process of redistribution of charge slows down.