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

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

Tardigrade Admin · Accepted Answer

Charge on 6 Î¼F capacitor = Charge on 12 Î¼F capacitor. ⇒ 6× 10- 6× 2 V=12× 10- 6 C ∴ Potential difference across 12 Î¼F capacitor is V=(12 × 10- 12/12 × 10- 12)=1 V Battery voltage =2 V+1 V=3 V

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

$4 V$

$6 V$

$9 V$

$3 V$

Charge on $6 μ F$ capacitor $=$ Charge on $12 μ F$ capacitor.
$\Rightarrow 6 \times 1 0^{- 6} \times 2 V = 12 \times 1 0^{- 6} C$
$∴$ Potential difference across $12 μ F$ capacitor is
$V = \frac{12 \times 1 0 ^{- 12}}{12 \times 1 0 ^{- 12}} = 1 V$
Battery voltage $= 2 V + 1 V = 3 V$

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

4V

6V

9V

3V