COMEDK · Physics · 18. Capacitance
A capacitor of capacitance \(4 \mu \mathrm{~F}\) is charged to a potential of 24 V and then connected in parallel to an uncharged capacitor of capacitance \(6 \mu \mathrm{~F}\). The final potential difference across each capacitor will be:
- A 9.6 V
- B 6.9 V
- C 8.2 V
- D 7.4 V
Answer & Solution
Correct Answer
(A) 9.6 V
Step-by-step Solution
Detailed explanation
The initial charge \(Q\) on the capacitor of capacitance \(C_1 = 4 \mu \text{F}\) charged to a potential \(V_1 = 24 \text{V}\) is given by \(Q = C_1 V_1\). \(Q = 4 \mu \text{F} \times 24 \text{V} = 96 \mu \text{C}\). When this capacitor is connected in parallel to an uncharged…
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