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(C) In a steady state,capacitors act as open circuits,so no current flows through the resistors. The potential difference across the entire capacitor

Vedclass · Accepted Answer

$V_{AF} = 28.5 	ext{ V}, V_{FB} = 71.4 	ext{ V}$

Vedclass · Answer

(C) In a steady state,capacitors act as open circuits,so no current flows through the resistors. The potential difference across the entire capacitor network ($C$ to $D$) is equal to the battery voltage,$V_{CD} = 100 	ext{ V}$.Looking at the circuit,the branch $E-F-G$ is in parallel with the branch $C-D$. However,the potential at $F$ is determined by the capacitive voltage divider formed by the branches connected between $C$ and $D$.The equivalent capacitance of the left branch (capacitors in series) is $C_1 = \frac{5 	imes 5}{5 + 5} = 2.5 	ext{ } \mu	ext{F}$.The equivalent capacitance of the right branch (capacitors in series) is $C_2 = \frac{2 	imes 2}{2 + 2} = 1 	ext{ } \mu	ext{F}$.Since these branches are in parallel across $100 	ext{ V}$,the potential at $F$ relative to $C$ and $D$ is determined by the series combination. The potential difference $V_{CF}$ is $100 	imes \frac{C_2}{C_1 + C_2} = 100 	imes \frac{1}{2.5 + 1} = 100 	imes \frac{1}{3.5} \approx 28.57 	ext{ V}$.Thus,$V_{AF} = V_{AC} + V_{CF} = 0 + 28.57 	ext{ V} = 28.57 	ext{ V}$.And $V_{FB} = V_{FD} + V_{DB} = (100 - 28.57) + 0 = 71.43 	ext{ V}$.Rounding to one decimal place,we get $V_{AF} = 28.5 	ext{ V}$ and $V_{FB} = 71.4 	ext{ V}$.

Find the potential difference between points $A$ and $F$,and $F$ and $B$.

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