Question

Capacitor of capacitance 10 micro farad is fully charged by 200V supply . The energy stored in the capacitor is

Answer 1

Given ,

Capacitance (C) = 10 µf or 10^(-5) f

Potential difference (v) = 200 v

We know that , the energy stored in a capacitor is given by

$\boxed{ \tt{Energy \: (U) = \frac{1}{2} {(v)}^{2}C}}$

Thus ,

$\tt : \mapsto U = \frac{1}{2} \times {(200)}^{2} \times {(10)}^{ - 5}$

$\tt : \mapsto U = \frac{40000 \times {(10)}^{ - 5} }{2}$

$\tt : \mapsto U =2 \times {(10)}^{ - 1}$

$\tt: \mapsto U =0.2 \: joule$

★ The energy stored in a capacitor is 0.2 J

Answer 2

Answer:

A capacitor of capacitance 10 microfarad is fully charged by a 200V supply. The energy stored in the capacitor is $0.2\ J$

Explanation:

• Let us consider a capacitor of capacitance C and with a potential difference between the capacitor plates V.
• The work done in charging a capacitor is stored as electric potential energy.

• Electric potential energy (E) stored in the capacitor   can be expressed as,

            $E=\frac{1}{2} C V^{2}$    ...(1)

In the question, it is given that,

$V= 200\ V$

$C =10\ \mu F = 10\times10^{-6} F$

Substitute these  values into equation (1)

We get,

$E=\frac{1}{2}\times 10\times10^{-6} \times 200^{2} = 0.2 \ J$

Thus,

The energy stored between the plates of the capacitor  will be   $0.2\ J$