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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 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
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The total charge is conserved, and the combined capacitance in parallel is the sum.
Updated On: Oct 1, 2026
- \(\frac{C_1V_1}{C_2}\)
- \(\frac{C_1V_1}{C_1+C_2}\)
- \(\frac{C_2V_1}{C_1}\)
- \(\frac{C_2V_1}{C_1+C_2}\)
Show Solution
The Correct Option is B
Solution and Explanation
Step 1: Use conservation.
Charge before $= C_1V_1 + 0$. Charge after $= C_1V_2 + C_2V_2$.
Step 2: Equate.
$C_1V_1 = (C_1 + C_2)V_2$.
Step 3: Solve.
$V_2 = \dfrac{C_1V_1}{C_1 + C_2}$. Since the denominator is larger than $C_1$, $V_2 < V_1$, as expected.
Final Answer:
Option (B). \[ \boxed{\frac{C_1V_1}{C_1+C_2}} \]
Charge before $= C_1V_1 + 0$. Charge after $= C_1V_2 + C_2V_2$.
Step 2: Equate.
$C_1V_1 = (C_1 + C_2)V_2$.
Step 3: Solve.
$V_2 = \dfrac{C_1V_1}{C_1 + C_2}$. Since the denominator is larger than $C_1$, $V_2 < V_1$, as expected.
Final Answer:
Option (B). \[ \boxed{\frac{C_1V_1}{C_1+C_2}} \]
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