Question

The two condensers of capacitances $2 \mu F$and $3 \mu F$ are in series. The outer plate of the first condenser is at 1000 volts and the outer plate of the second condenser is earthed. The potential of the inner plate of each condenser is

• A. 300 volts
• B. 500 volts
• C. 600 volts
• D. 400 volts

Answer 1

Answer: (D)

400 volts

Answer 2

Here, potential difference across the combination is

$V _ { A } - V _ { B } = 1000 V$

Equivalent capacitance $C _ { e q } = \frac { 2 \times 3 } { 2 + 3 } = \frac { 6 } { 5 } \mu F$

Hence, charge on each capacitor will be

$Q = C _ { e q } \times \left( V _ { A } - V _ { B } \right)$

So potential difference between A and C,

$V _ { A } - V _ { C } = \frac { 1200 } { 2 } = 600 \mathrm {~V}$ ⇒ $1000 - V _ { C } = 600$

⇒ $V _ { c } = 400 \mathrm {~V}$