Question

Find potential difference between points $A\&F$ and $F\&B$

• A. $V_{AF}=10.2V{V}_{FB}=15.4V$
• B. $V_{AF}=22.3V{V}_{FB}=28.9V$
• C. $V_{AF}=28.5V{V}_{FB}=71.4V$
• D. $V_{AF}=42.1V{V}_{FB}=53.1V$

Answer 1

Answer: (C)

The given circuit is

The two capacitors of $5\mu F$ each are in parallel. So, their equivalent capacitance is $(5+5)10\mu F$. Similarly, two $2\mu F$ capacitors are also in parallel. So, their equivalent capacitance is $(2+2)4\mu F$. The circuit is now reduced to as shown

In steady state, when all the capacitors are charged, no current flow in the circuit. So, potential across resistors is zero.
As in parallel combination, potential is same in both wires $CD$ and $BG$, i.e. $100V$.
In series combination, the potential is divided between the capacitor in the inverse ratio of their capacitance,
i.e. $\frac{V_{AP}}{V_{FB}}=\frac{4}{10}=\frac{2}{5}$
or $2:5\therefore V_{AF}=\frac{2}{7}\times 100=28.5V$
and $V_{FB}=\frac{5}{7}\times 100=71.4V$