1.

a capacitor of capacitance value 1 microfarad is charged to 30 volts and the battery is then disconnected if it is connected across a 2 microfarad capacitor then energy lost by system

Answer» ENERGY LOSS =
\frac{1}{2} \frac{c1c2}{c1 + c2} (v1 - v2) {}^{2}
c1 = 10^-6 F
c2 = 2 * 10^-6 F
v1 = o v
v2 = 30v

:. E =
\frac{1}{2} \frac{1 \times 10 {}^{ - 6} \times 2 \times 10 {}^{ - 6} }{1 \times 10 {}^{ - 6} + 2 \times 10 {}^{ - 6} } (0 - 30) {}^{2}
:. E =
\frac{1}{2} \frac{2 \times 10 {}^{ - 12} }{3 \times 10 {}^{ -6} } \times 9 \times 10 {}^{2}
:. E =
3 \times 10 {}^{ - 4} j


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