Question

A \(5\,\mu F\) capacitor is fully charged by a \(12\,V\) battery and then disconnected. If it is connected now in parallel to an uncharged capacitor, the voltage across it is \(3\,V\). Then the capacity of the uncharged capacitor is

• A. \(5\,\mu F\)
• B. \(15\,\mu F\)
• C. \(50\,\mu F\)
• D. \(10\,\mu F\)
• E. \(25\,\mu F\)

Hint:

Charge remains conserved when capacitors are disconnected from battery.

Answer 1

The Correct Option is B

Concept: When capacitors are connected, total charge is conserved.

Step 1: Initial charge on capacitor. \[ Q = CV = 5 \times 12 = 60\,\mu C \]

Step 2: After connection, voltage becomes 3 V. Total capacitance = \(5 + C\)

Step 3: Apply charge conservation. \[ Q = (5 + C)\times 3 \]

Step 4: Substitute. \[ 60 = 3(5 + C) \Rightarrow 60 = 15 + 3C \]

Step 5: Solve. \[ 3C = 45 \Rightarrow C = 15\,\mu F \]

Step 6: Conclusion. \[ \boxed{15\,\mu F} \]