A 400 pF capacitor is charged with a 100 V battery. After disconnecting the battery this capacitor is connected with another 400 pF capacitor. Then the energy loss is:
\( 1 \times 10^{-6}\text{ J} \)
\( 2 \times 10^{-6}\text{ J} \)
\( 3 \times 10^{-6}\text{ J} \)
\( 4 \times 10^{-6}\text{ J} \)
Solution:
Energy loss when connecting two capacitors is \( \Delta U = \frac{1}{2} \frac{C_1 C_2}{C_1 + C_2} (V_1 - V_2)^2 \). Here, \( C_1 = C_2 = 400\text{ pF} \), \( V_1 = 100\text{ V} \), and \( V_2 = 0 \). This gives \( \Delta U = \frac{1}{4} C V^2 = 1 \times 10^{-6}\text{ J} \).
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