Question

A capacitor of capacitance \(4\,\mu F\) is charged to a potential of \(24\,V\) and then connected in parallel to an uncharged capacitor of capacitance \(6\,\mu F\). The final potential difference across each capacitor will be:

• A. \(6.9\,V\)
• B. \(8.2\,V\)
• C. \(9.6\,V\)
• D. \(7.4\,V\)

Hint:

In capacitor connection problems, charge is conserved but energy may not be conserved. Always use \(Q_{\text{initial}} = Q_{\text{final}}\).

Answer 1

The Correct Option is C

Step 1: Find initial charge on the charged capacitor.
\[ Q = CV \]
\[ Q = 4\,\mu F \times 24\,V \]
\[ Q = 96\,\mu C \]

Step 2: Understand the process.
When connected in parallel, total charge is conserved and redistributes over both capacitors.

Step 3: Find total capacitance.
\[ C_{\text{total}} = 4 + 6 = 10\,\mu F \]

Step 4: Find final common voltage.
\[ V_f = \frac{Q_{\text{total}}}{C_{\text{total}}} \]
\[ V_f = \frac{96}{10} \]
\[ V_f = 9.6\,V \]

Step 5: Interpretation.
Both capacitors will have the same final voltage because they are connected in parallel.

Step 6: Energy consideration (optional).
Some energy is lost during redistribution, but charge is conserved.

Step 7: Final answer.
\[ \boxed{9.6\,V} \]