Question

Three capacitors of 2μ f, 3μ f and 6μ f are joined in series and the combination is charged by means of a 24 volt battery. The potential difference between the plates of the 6μ f capacitor is

• A. 4 volts
• B. 6 volts
• C. 8 volts
• D. 10 volts

Answer 1

Answer: (A)

4 volts

Answer 2

Equivalent capacitance of the network is $\frac{1}{C_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6}$

$C_{eq} = 1\mu F$

Charge supplied by battery Q = C_eq.V ⇒ 1 × 24 = 24 μC

Hence potential difference across 6μ F capacitor $= \frac{24}{6} = 4volt.$