Question

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 1

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 =
$3 \times 10 {}^{ - 4} j$

Answer 2

Answer:150micro joules

Explanation: