Question

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

• A. during charging, more heat is produced in A than in B
• B. in the steady state, heat is produced at the same rate in A and B
• C. in the steady state, energy stored in C is $\frac{1}{4}C{ϵ}^2$
• D. in the steady state, energy stored in C is $\frac{1}{8}C{ϵ}^2$

Answer 1

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

The correct options are
A during charging, more heat is produced in A than in B
B in the steady state, heat is produced at the same rate in A and B
D in the steady state, energy stored in C is $\frac{1}{8}C{ϵ}^2$

When S is closed, the capacitor will start charging through resistor A and less current will pass through B. So during charging more heat will produce on A than B.
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.
As both A and B have equal resistance so in steady state voltage drop of each resistor is $ϵ/2$.
Here voltage across C is the voltage across B i.e $V_C=\frac{ϵ}{2}$
Energy stored, $U=\frac{1}{2}C{V}_C^2=\frac{1}{8}C{ϵ}^2$