Question

Two capacitors, $C_1$ and $C_2$, are connected in series with a 12.0-volt direct-current source. The capacitance of $C_1$ is $600F$. When a voltmeter is connected across $C_1$, the voltmeter reads 3.0 volts.
What is the capacitance of $C_2$?

• A. $200F$
• B. $300F$
• C. $600F$
• D. $900F$
• E. $1,800F$

Answer 1

Answer: (A)

$200F$

As both the capacitors are connected in series, thus both must have same amount of charge.

Given : $C_1=600$ F $V_1=3.0$ volts

$\therefore$ Charge on capacitor $C_1$, $Q=C_1{V}_1=600\times 3.0=1800$ coulombs

Potential drop across $C_2$, $V_2=12.0-3.0=9.0$ volts

$\therefore$ Capacitance of capacitor $C_2$, $C_2=\frac{Q}{V_2}=\frac{1800}{9.0}=200$ F