Question

In the given circuit, capacitor $C$ is charged by cell of emf $E$ and reaches a steady state. At $t=0$, switch $S$ is closed and $A$ and $B$ are equal resistances. Now, consider the following statements.


(1) In steady state, rate of heat production in $A$ and $B$ is different.
(2) During charging, more heat is produced in $B$ than in $A$
(3) In steady state, rate of heat production is same in $A$ and $B$
(4) In steady state, energy stored in capacitor is $\frac{1}{8}C{E}^2$,
Which of the above statements are correct?

• A. $1,3,4$
• B. $1,2$
• C. $2,3,4$
• D. $1,2,4$

Answer 1

The correct option is C $2,3,4$

When $S$ is closed, the capacitor will start charging through $B$ and less current will pass through $A$. So, during charging more heat will be produced in $B$ than $A$.

In steady state, a constant current will flow through whole circuit. Thus, heat is produced at the same rate in $A$ and $B$ (in steady state).

In steady state, capacitor will be open circuit. So, potential drop across $A$ in steady state = potential drop across $B$ in steady state = potential drop across capacitor in steady state $=\frac{E}{2}$.

Hence, in steady state energy stored in capacitor
$=\frac{1}{2}C{V}^2=\frac{1}{2}C{\left(\frac{E}{2}\right)}^2$
$=\frac{1}{8}C{E}^2$