ExamBro
ExamBro
MHT CET · Physics · Capacitance

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

  1. A \(\frac{\mathrm{V}_1 \mathrm{C}_2}{\mathrm{C}_1}\)
  2. B \(\frac{\mathrm{C}_2}{\mathrm{C}_1+\mathrm{C}_2}\)
  3. C \(\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_2}\)
  4. D \(\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_1+\mathrm{C}_2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_1+\mathrm{C}_2}\)

Step-by-step Solution

Detailed explanation

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, \(\mathrm{V}_2=\frac{\mathrm{Q}}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{C}_1 \mathrm{~V}_1}{\mathrm{C}_1+\mathrm{C}_2}\)
From MHT CET
Explore more questions on app