Question

A capacitor of capacity C1 is charged upto V volts and then connected to an uncharged capacitor of capacity C2. The final potential difference across each will be: एक संधरित्र, जिसकी धारिता C1 है, को V वोल्ट से आवेशित किया जाता है तथा उसको एक अनआवेशित संधारित्र, जिसकी धारिता C2 है, से जोड़ दिया जाता है। तब अन्तिम विभवान्तर दोनों संधारित्रों के परितः होगाः-

• A. $\frac{C_2V}{C_1+C_2}$
• B. $\frac{C_1V}{C_1+C_2}$
• C. $(1+\frac{C_2}{C_1})V$
• D. $(1-\frac{C_2}{C_1})V$

Answer 1

Answer: (B)

$\frac{C_1V}{C_1+C_2}$

Answer 2

Explanation of the solution:

1. Initial State:

   • Capacitor $C_1$ is charged to $V$ volts. The initial charge on $C_1$ is $Q_1 = C_1V$.
   • Capacitor $C_2$ is uncharged, so its initial charge is $Q_2 = 0$.
   • The total initial charge in the system is $Q_{initial} = Q_1 + Q_2 = C_1V + 0 = C_1V$.

2. Final State (After Connection):

   • When the two capacitors are connected, they form a parallel combination. Charge redistributes between them until they reach a common final potential difference, let's call it $V_f$.
   • The final charge on $C_1$ will be $Q'_1 = C_1V_f$.
   • The final charge on $C_2$ will be $Q'_2 = C_2V_f$.
   • The total final charge in the system is $Q_{final} = Q'_1 + Q'_2 = C_1V_f + C_2V_f = (C_1 + C_2)V_f$.

3. Conservation of Charge:

   • According to the principle of conservation of charge, the total charge in an isolated system remains constant. Therefore, the total initial charge must be equal to the total final charge.
   • $Q_{initial} = Q_{final}$
   • $C_1V = (C_1 + C_2)V_f$

4. Final Potential Difference:

   • Solving for $V_f$: $V_f = \frac{C_1V}{C_1+C_2}$

This is the final potential difference across each capacitor.