Question

A condenser of capacity \(C_1\) is charged to potential \(V_1\) and then disconnected. Uncharged capacitor of capacity \(C_2\) is connected in parallel with \(C_1\). The resultant potential \(V_2\) is

• A. \(\frac{C_1V_1}{C_2}\)
• B. \(\frac{C_1V_1}{C_1+C_2}\)
• C. \(\frac{C_2V_1}{C_1}\)
• D. \(\frac{C_2V_1}{C_1+C_2}\)

Hint:

The total charge is conserved, and the combined capacitance in parallel is the sum.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When the charged capacitor, already disconnected from its battery, is connected in parallel with an uncharged one, charge flows until both have the same potential. The total charge does not change.

Step 2: Initial charge.
\(Q = C_1V_1\).

Step 3: Final state.
The capacitors in parallel have capacitance \(C_1 + C_2\) and common potential \(V_2\), so
\[ V_2 = \frac{Q}{C_1 + C_2} = \frac{C_1V_1}{C_1 + C_2} \]

Step 4: Check the options.
Options (A), (C) and (D) divide by the wrong capacitance or use the wrong charge.

Final Answer:
The common potential is \(\dfrac{C_1V_1}{C_1 + C_2}\), option (B). \[ \boxed{\frac{C_1V_1}{C_1+C_2}} \]