MHT CET · Physics · Electrostatics
If a capacitor of capacity \(900 \mu \mathrm{F}\) is charged to \(100 \mathrm{~V}\) and its total energy is transferred to a capacitor of capacity \(100 \mu \mathrm{F}\), then its potential will be
- A \(30 \mathrm{~V}\)
- B \(200 \mathrm{~V}\)
- C \(300 \mathrm{~V}\)
- D \(400 \mathrm{~V}\)
Answer & Solution
Correct Answer
(C) \(300 \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Let the unknown potential be \(\mathrm{V}_{\mathrm{u}}\)
Energy of a capacitor \(=\frac{1}{2} \mathrm{CV}^2\)
\(\Rightarrow \frac{1}{2} C_1 V_1^2=\frac{1}{2} C_2\left(V_u\right)^2 \)
\( \frac{1}{2} \times 900 \times 10^{-6} \times(100)^2=\frac{1}{2} \times 100 \times 10^{-6} \times\left(V_u\right)^2 \)
\( \left(V_u\right)^2=9 \times 100^2 \)
\( \therefore V_u=300 \mathrm{~V}\)
Energy of a capacitor \(=\frac{1}{2} \mathrm{CV}^2\)
\(\Rightarrow \frac{1}{2} C_1 V_1^2=\frac{1}{2} C_2\left(V_u\right)^2 \)
\( \frac{1}{2} \times 900 \times 10^{-6} \times(100)^2=\frac{1}{2} \times 100 \times 10^{-6} \times\left(V_u\right)^2 \)
\( \left(V_u\right)^2=9 \times 100^2 \)
\( \therefore V_u=300 \mathrm{~V}\)
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