A $400 p F$ capacitor is charged with a $100 V$ battery. After disconnecting battery this capacitor is connected with another $400 p F$ capacitor. Then find out energy loss.

1 \times 10^{- 6} J

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 \times 10^{- 6} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 \times 10^{- 6} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 \times 10^{- 6} J

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $1 \times 10^{- 6} J$
$\begin{matrix} L o s s o f P . E . = \frac{1}{2} \times \frac{c_{1} c_{2}}{c_{1} + c_{2}} {(v_{1} - v_{2})}^{2} = \frac{1}{2} \cdot \frac{400}{2} P F {(100 - 0)}^{2} = 10^{- 6} J \end{matrix}$
Ans. (A)

Suggest Corrections

Similar questions

Q. A

400 p F

capacitor is charged with a

100 V

battery. After disconnecting battery this capacitor is connected with another

400 p F

capacitor. Then find out energy loss.

Q. A capacitor is charged with a battery and energy stored is

U

. After disconnecting battery another capacitor is of same capacity is connected in parallel with it. Then energy stored in each capacitor is:

Q. A capacitor is charged with a battery and energy stored is

U

. After disconnecting the battery, another identical uncharged capacitor is connected in parallel with it. The energy stored in each capacitor is now

A $4 μ F$ capacitor is charged by a 200V battery. It is then disconnected from the supply and is connected to another uncharged $4 μ F$ capacitor. During this process, Loss of energy (in J) is :

Q. The energy of a charged capacitor is

U

. Another identical capacitor is connected parallel to the first capacitor, after disconnecting the battery. The total energy of the system of these capacitors will be

Join BYJU'S Learning Program

A 400pF capacitor is charged with a 100V battery. After disconnecting battery this capacitor is connected with another 400pF capacitor. Then find out energy loss.

The correct option is A 1×10−6JLossofP.E.=12×c1c2c1+c2(v1−v2)2=12⋅4002PF(100−0)2=10−6JAns. (A)

A $400 p F$ capacitor is charged with a $100 V$ battery. After disconnecting battery this capacitor is connected with another $400 p F$ capacitor. Then find out energy loss.

The correct option is A $1 \times 10^{- 6} J$
$\begin{matrix} L o s s o f P . E . = \frac{1}{2} \times \frac{c_{1} c_{2}}{c_{1} + c_{2}} {(v_{1} - v_{2})}^{2} = \frac{1}{2} \cdot \frac{400}{2} P F {(100 - 0)}^{2} = 10^{- 6} J \end{matrix}$
Ans. (A)