Question

Two capacitors marked $(12 \, \mu F \, 200 \, V) (24 \, \mu F \, 200 \, V)$ are connected in series. The combination acts as a capacitor with breakdown voltage

• A. $( 36 \, \mu F \, 200 \, V )$
• B. $( 8 \, \mu F \, 400 \, V )$
• C. $(8 \, \mu F \, 300 \, V )$
• D. $( 36 \, \mu F \, 400 \, V )$

Answer 1

Answer: (C)

Limiting charge on capacitor of $12\, \mu F$
$q_1 = 12 \, \mu F \times 200 \, V = 2400 \, \mu C$
Limiting charge on capacitor of $24 \, \mu F$,
$q_2 = 24 \, \mu F \times 200 \, V = 4800 \, \mu C$
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
$V_b = \frac{Q}{C} = \frac{2400 \, \mu C}{8 \mu F} = 300 \, V$