Question

A 15-µF capacitor and a 25-µF capacitor are connected in parallel, and charged to a potential difference of 60 V. How much energy is then stored in this capacitor combination?

Answer 1

Answer:

Energy stored, E = 0.072 J

Explanation:

Given that,

Capacitance, $C_1=15\ \mu F$

Capacitance, $C_2=25\ \mu F$

These two capacitor are connected in parallel, and charged to a potential difference of, V = 60 volts

We know that in parallel combination of capacitor, the equivalent capacitance is given by :

$C=C_1+C_2\\\\C=(15+25)\ \mu F\\\\C=40\times 10^{-6}\ F$

The energy stored in the capacitor is given by :

$E=\dfrac{1}{2}CV^2\\\\E=\dfrac{1}{2}\times 40\times 10^{-6}\times (60)^2\\\\E=0.072\ J$

So, the energy stored in the capacitor in this capacitor combination is 0.072 J.