Question

A 25 µF capacitor is connected to a 12-volt battery. How much energy can be stored in the capacitor? (1F = 106 µF)

Answer 1

w=1/2cvsqr....1/2*25*10^-6*144
0.0000125*144=0.0018=18micro farad

Answer 2

Answer:

Energy stored in the capacitor is 0.0018 joules.

Explanation:

It is given that,

Capacitance of the capacitor, $C=25\ \mu F=25\times 10^{-6}\ F$

Voltage of the battery, V = 12 volt

We have to find the energy stored in the capacitor. The energy stored inside the capacitor is given by :

$E=\dfrac{1}{2}CV^2$

$E=\dfrac{1}{2}\times 25\times 10^{-6}\ F\times (12\ V)^2$

E = 0.0018 Joules

Hence, the energy stored in the capacitor is 0.0018 Joules.