Question

A capacitor of capacitance 50 micro farads is charged to a potential difference of 200 Volts. Find out energy stored in the condenser

Answer 1

Explanation:

Energy stored in the capacitor is given by

E=1/2CV^2

=1/2*50*10^-6*(200)^2

=1 farad volt^2

= 1 J

Answer 2

Question:

A capacitor of capacitance 50 micro farads is charged to a potential difference of 200 Volts. Find out energy stored in the condenser.

Formula used :

Energy stored in a capacitor ,

${\purple{\boxed{\large{\bold{U=\frac{Q{}^{2}}{2C}=\frac{1}{2} CV{}^{2}}}}}}$

Solution :

⇒Given :

Capacitance,C= 5OμF =$50\times10{}^{-6}$F

Potential difference,V=200volts

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

Now put the values

$U= \frac{1}{2} \times 50 \times 10 {}^{ - 6} \times( 200) {}^{2}$

$U= \frac{1}{2} \times 50 \times 10 {}^{ - 6} \times 4 \times 10 {}^{4}$

$U= 50 \times 10 {}^{ - 6} \times 10 {}^{4} \times 2$

$U= 100 \times 10 {}^{ - 6 + 4}$

$\implies u = 1J$

Therefore,the Energy stored in the condenser is 1 J.