Question

In the circuit, shown in figure, the potential difference between the $3$ points $A$ and $B$ in the steady state is :

• A. zero
• B. $6V$
• C. $4V$
• D. $\frac{10}{3}V$

Answer 1

The correct option is D $\frac{10}{3}V$

In steady state, no current flow through the resistors. So the corresponding circuit is as shown in figure.
here, $V_{AC}=10V$
Net capacitance $C_{AB}=3||3=3+3=6\mu F$
Net capacitance $C_{BC}=1||2=1+2=3\mu F$
Now $C_{AB}$ and $C_{BC}$ in series, so $C_{AC}=\frac{6\times 3}{6+3}=2\mu F$
Charge, $Q_{AC}=C_{AC}{V}_{AC}=2\times 10=20\mu C$
As $C_{AB}$ and $C_{BC}$ in series, so charge on each equal to $Q_{AC}$
thus, $V_{AB}=\frac{Q_{AC}}{C_{AB}}=\frac{20}{6}=\frac{10}{3}V$