Search any question & find its solution
Question:
Answered & Verified by Expert
A condenser of capacity ' $\mathrm{C}_1$ ' is charged to potential ' $\mathrm{V}_1$ ' and then disconnected. Uncharged capacitor of capacity ' $\mathrm{C}_2$ ' is connected in parallel with ' $\mathrm{C}_1$ '. The resultant potential ' $\mathrm{V}_2$ ' is
Options:
Solution:
2722 Upvotes
Verified Answer
The correct answer is:
$\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_1+\mathrm{C}_2}$
The charge stored by the charged condenser is
$$
\mathrm{Q}=\mathrm{C}_1 \mathrm{~V}_1
$$
When the uncharged condenser is connected in parallel, the effective capacitance becomes $\left(\mathrm{C}_1+\mathrm{C}_2\right)$
Hence the potential, $\mathrm{V}_2=\frac{\mathrm{Q}}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_1+\mathrm{C}_2}$
$$
\mathrm{Q}=\mathrm{C}_1 \mathrm{~V}_1
$$
When the uncharged condenser is connected in parallel, the effective capacitance becomes $\left(\mathrm{C}_1+\mathrm{C}_2\right)$
Hence the potential, $\mathrm{V}_2=\frac{\mathrm{Q}}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_1+\mathrm{C}_2}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.