Question

Two capacitors marked $(12\mu F200V)(24\mu F200V)$ are connected in series. The combination acts as a capacitor with breakdown voltage

• A. $(36\mu F200V)$
• B. $(8\mu F400V)$
• C. $(8\mu F300V)$
• D. $(36\mu F400V)$

Answer 1

Answer: (C)

Limiting charge on capacitor of $12\mu F$
$q_1=12\mu F\times 200V=2400\mu C$
Limiting charge on capacitor of $24\mu F$,
$q_2=24\mu F\times 200V=4800\mu C$
Now, these two capacitors are connected in series so equivalent capacitance
$C=\frac{C_1{C}_2}{C_1+C_2}=\frac{12\times 24}{12+24}=8\mu F$
Maximum charge that can flow through the capacitors is $2400\mu C$.
So, breakdown voltage of the circuit
$V_b=\frac{Q}{C}=\frac{2400\mu C}{8\mu F}=300V$