Question

Three capacitors $C_1$, $C_2$ and $C_3$ are connected as shown in the figure to a battery of V volt. If the capacitor $C_3$ breaks down electrically the change in total charge on the combination of capacitors is:

• A. $(C_1+C_2)V[1-C_3/(C_1+C_2+C_3)]$
• B. $(C_1+C_2)V[1-(C_1+C_2)/(C_1+C_2+C_3)]$
• C. $(C_1+C_2)V[1+C_3/(C_1+C_2+C_3)]$
• D. $(C_1+C_2)V[1-C_2/(C_1+C_2+C_3)]$

Answer 1

The correct option is A $(C_1+C_2)V[1-C_3/(C_1+C_2+C_3)]$

Equivalent capacitance of circuit,

$\frac{1}{C_{eq}}=\frac{1}{C_3}+\frac{1}{C_1+C_2}$

since $C_1$ and $C_2$ are in parallel and which is in series with $C_3$

$\frac{1}{C_{eq}}=\frac{C_1+C_2+C_3}{C_3(C_1+C_2)}$

Since V is the voltage of battery,charge $q=C_{eq}V=\frac{C_3(C_1+C_2)}{C_1+C_2+C_3}V$

If the capacitor $C_3$ breaks down, then effective capacitance,

$C_{eq}=C_1+C_2$

New Charge

$q^{{}^{\prime }}=C_{eq}^{{}^{\prime }}V=(C_1+C_2)V$

Change in total charge $q^{{}^{\prime }}-q$

$(C_1+C_2)V-\frac{C_3(C_1+C_2)}{C_1+C_2+C_3}V$

$(C_1+C_2)V[1-\frac{C_3}{C_1+C_2+C_3}]$