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KCET · Physics · Motion In One Dimension

Two capacitors of \( 10 \mathrm{pF} \) and \( 20 \mathrm{pF} \) are connected to \( 200 \mathrm{~V} \) and \( 100 \mathrm{~V} \) sources respectively. If
they are connected by the wire, what is the common potential of the capacitors ?

  1. A \( 133.3 \) volt
  2. B \( 150 \) volt
  3. C \( 300 \) volt
  4. D \( 400 \) volt
Verified Solution

Answer & Solution

Correct Answer

(A) \( 133.3 \) volt

Step-by-step Solution

Detailed explanation

Common potential, \( V=\frac{C V_{1}+C_{2} V_{2}}{C_{1}+C_{2}} \)
\( =\frac{10 \times 10^{-12} \times 200+20 \times 10^{-11} \times 100}{10 \times 10^{-12}+20 \times 10^{-12}} \)
\( \therefore V=\frac{10^{-12} \times 200(1+1)}{10^{-12} \times 10(1+2)}=\frac{200 \times 2}{3}=133.33 \mathrm{~V} \)
Thus, common potential of the capacitors is \( 133.33 \mathrm{~V} \).