Question

O A capacitor of capacitance
2nF is changed
d to a
potential of lov. Calculate
stored
energy​

Answer 1

Explanation:

Initial energy stored in capacitor 2 μF:U

i

=

2

1

2(V)

2

=V

2

Final voltage after switch 2 is on: V

f

=

C

1

+C

2

C

1

V

1

=

10

2V

=0.2 V

Final energy in both the capacitors, U

f

=

2

1

(C

1

+C

2

)V

f

2

=

2

1

×10×(

10

2V

)

2

=0.2 V

2

Therefore, energy dissipated =

V

2

V

2

−0.2V

2

×100=80%

Answer 2

Answer:

A capacitor of capacitance  2nF is changed to a  potential of 10v. The  stored

energy will be $10^{-7} \ J$

Explanation:

• 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)

Where C - capacitance of the capacitor

            V - potential difference between the capacitor plates

In the question, it is given that,

$V= 10\ V$

$C =2\ \n F = 2\times10^{-9} F$

Substitute these  values into equation (1)

We get,

$E=\frac{1}{2}\times 2\times10^{-9} \times 10^{2} = 10^-7\ \ J$

Thus,

The energy stored between the plates of the capacitor  will be   $10^{-7} \ J$