Question

Two capacitors of capacitance $6 \, μF$ and $12 \, μF$ are connected in series with a battery. The voltage across the $6 \, μF$ capacitor is $2 \, V$ . The total battery voltage is

• A. $4V$
• B. $6V$
• C. $9V$
• D. $3V$

Answer 1

Answer: (D)

Charge on $6 \, μF$ capacitor $=$ Charge on $12 \, μF$ capacitor.
$\Rightarrow 6\times 10^{- 6}\times 2 \, V=12\times 10^{- 6} \, C$
$\therefore $ Potential difference across $12 \, μF$ capacitor is
$V=\frac{12 \times 10^{- 12}}{12 \times 10^{- 12}}=1 \, V$
Battery voltage $=2 \, V+1 \, V=3 \, V$