Question

The energy required to charge a capacitor of
Su by connecting D.C. source of 20 kV is
(a) 10 kj
(b) 5 kJ
(c) 2 kJ
(d) 1 kJ​

Answer 1

Given:

Capacitance $=C=$ 5 μF

Voltage $=V=$ 20 kV

To find:

We need to find the energy required to charge the capacitor.

Solution:

We know that, energy of the capacitor is given by the formula,

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

Where, $E$ is the energy of the capacitor, $C$ is the capacitance and $V$ is the voltage.

Now substitute the given values in the above formula to get the value of energy.

So,

$\Rightarrow E=\frac{1}{2} CV^2$

Substituting the given values,

$\Rightarrow E=\frac{1}{2} (5 \ \mu\text{F})(20\text{ kV})^2$

$\Rightarrow E=\frac{1}{2} (5\times 10^{-6}\ \text{F})(20\times 10^3\ \text{V})^2$

Expanding the terms,

$\Rightarrow E=\frac{1}{2} (5\times 10^{-6}\ \text{F})(20\times 20\times 10^3\times 10^3 \ \text{V})$

Simplifying the terms,

$\Rightarrow E=\frac{1}{2} (5\times20\times 20)$

$\Rightarrow E=(5\times10\times 20)$

$\Rightarrow E=1000$

$\Rightarrow E=1 \text{ kJ}$

Therefore, we need 1kJ to charge the capacitor.