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A capacitor \(\mathrm{C}_1\) is charged to a potential difference \(\mathrm{V}\). The charging battery is then removed and the capacitor is connected to an uncharged capacitor \(\mathrm{C}_2\). The potential difference across the combination is
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The correct answer is:
\(\frac{\mathrm{VC}_1}{\left(\mathrm{C}_1+\mathrm{C}_2\right)}\)
Charge \(\mathrm{Q}=\mathrm{C}_1 \mathrm{~V}\)
Total capacity of combination (parallel)
\(\mathrm{C}=\mathrm{C}_1+\mathrm{C}_2\)
\(\text {P.D. }=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{\mathrm{C}_1 \mathrm{~V}}{\mathrm{C}_1+\mathrm{C}_2}\)
Total capacity of combination (parallel)
\(\mathrm{C}=\mathrm{C}_1+\mathrm{C}_2\)
\(\text {P.D. }=\frac{\mathrm{Q}}{\mathrm{C}}=\frac{\mathrm{C}_1 \mathrm{~V}}{\mathrm{C}_1+\mathrm{C}_2}\)
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