Question

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

• A. During charging, more heat is produced in A then in B
• B. In steady state, heat is produced at the same 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

Answer: (A), (B), (D)

The correct options are
A During charging, more heat is produced in A then in B
B In steady state, heat is produced at the same rate in A and B
D In the steady state, energy stored in C is $\frac{1}{2}C{\epsilon }^2$

During charging the current through A is the main current passing through the battery. So, this current is more than the current passing through B, i.e $I=I_A=I_C+I_B$. Hence during charging more heat is produced in A. $(H=I^{2}Rt)$
In steady state , no current flow in the capacitor wire so $I_C=0$
so, $I_A=I_B$ ,hence heat is produced at the same rate in A and B.
In steady state , the energy stored in capacitor is $U=\frac{1}{2}C{\epsilon }^2$
ans : A, B