Question

A \(1\,\mu F\) capacitor is charged to \(12\,V\) and then connected to an identical uncharged capacitor. What will be the common potential?

• A. \(12\,V\)
• B. \(6\,V\)
• C. \(3\,V\)
• D. \(4\,V\)

Hint:

For two identical capacitors, one charged and the other uncharged, \[ \boxed{V_f=\frac{V}{2}} \] because the total charge remains constant while the total capacitance doubles. Always apply the principle of conservation of charge in capacitor-sharing problems.

Answer 1

The Correct Option is B

The initial charge on the capacitor is $Q = CV = 1 \times 12 = 12\,\mu$C. When connected to an identical uncharged capacitor, total charge is conserved and distributes equally between the two, so the final voltage is $V_f = Q/(2C) = 6$ V.