Question

In the circuit shown, $A$ and $B$ are equal resistances. When $S$ is closed, the capacitor $C$ charge from the cell of emf $\epsilon$ reaches a steady state. The correct option is

• A. During charging, more heat is produced in $A$ than in $B$
• B. In the steady state, heat is produced at different rate in $A$ and $B$
• C. In the steady state, energy stored in $C$ is $\frac{1}{4}C{\epsilon }^2$
• D. In the steady state, energy stored in $C$ is $\frac{1}{2}C{\epsilon }^2$

Answer 1

The correct option is A During charging, more heat is produced in $A$ than in $B$

In steady state potential difference across $C$ is $\frac{\epsilon }{2}$
$\therefore$ Energy stored in $C$ is $=\frac{1}{2}C{\left(\frac{\epsilon }{2}\right)}^2=\frac{C{\epsilon }^2}{8}$
$\therefore$ (c) and (d) are incorrect.
In steady state current through resistors $A$ and $B$ will be same. So, heat will be produced at the same rate in both $A$ and $B$ as their resistances are same.
$\therefore$ (b) is also incorrect.
Before the current reaches the steady state, current in $A$ is more than that in $B$. So the heat dissipated in $A$ is more than that in $B$.