Question

Question of capacitor and capacitance:-​

Answer 1

Que:- A capacitor of capacitance $C_1$ is charged by connecting it to a battery . The battery is now removed and and this capacitor is connected to a second uncharged capacitor of capacitance $C_2$ . If the charge distributes equally on the two capacitors , the ratio of total energy stored in the capacitors to the total energy stored in them before the connection is :-

Soln:-

$\green{Given}$

• Two capacitors of capacitance $C_1$ and $C_2$

• $C_1$ is charged by a battery.When the battery is removed and it is connected to an another capacitor, charge is distributed equally.

$\green{Formula}$

• Q = CV

• Energy stored in a capacitor is given by $\frac{1}{2} C {V}^{2} = \frac{ {Q}^{2} }{2C}$

$\green{Concept}$

• When the battery is removed , the charge on the capacitor remains constant.

$\green{Solution}$

Let the charge on the capacitor $C_1$ be Q when it is connected across the battery. It would be same when the battery is removed.

Now, A/Q When it connected across $C_2$, Charge on both is same.

Hence, charge on each capacitor will be same that is $\frac{Q}{2}$

Now, voltage across them will be same.

$\frac{Q}{2} c_1$ = $\frac{Q}{2} c_2$

=> $C_1$ = $C_2$

Initial energy stored on the capacitor $C_1$ = $\frac{ {Q}^{2} }{2C_1}$

Final energy on the capacitor when charge has been halved = $\frac{ \frac{Q}{2}^2 }{2C_1}$ = $\frac{ {Q}^{2} }{4C_1}$

Final energy on the capacitor $C_2$ = $\frac{ \frac{Q}{2}^2 }{2C_2}$ = $\frac{ \frac{Q}{2}^2 }{2C_1}$ ( Since, $C_1$ = $C_2$ )

Hence, Total final energy = $\frac{ {Q}^{2} }{2C_1}$ = Total initial energy.

$\boxed { \pink { Hence,\:the\:ratio\:is\:1.} }$

Answer 2

Answer:

Que:- A capacitor of capacitance C_1C

1

is charged by connecting it to a battery . The battery is now removed and and this capacitor is connected to a second uncharged capacitor of capacitance C_2C

2

. If the charge distributes equally on the two capacitors , the ratio of total energy stored in the capacitors to the total energy stored in them before the connection is :-

Soln:-

\green{Given}Given

Two capacitors of capacitance C_1C

1

and C_2C

2

C_1C

1

is charged by a battery.When the battery is removed and it is connected to an another capacitor, charge is distributed equally.

\green{Formula}Formula

Q = CV

Energy stored in a capacitor is given by \frac{1}{2} C {V}^{2} = \frac{ {Q}^{2} }{2C}

2

1

CV

2

=

2C

Q

2

\green{Concept}Concept

When the battery is removed , the charge on the capacitor remains constant.

\green{Solution}Solution

Let the charge on the capacitor C_1C

1

be Q when it is connected across the battery. It would be same when the battery is removed.

Now, A/Q When it connected across C_2C

2

, Charge on both is same.

Hence, charge on each capacitor will be same that is \frac{Q}{2}

2

Q

Now, voltage across them will be same.

\frac{Q}{2} c_1

2

Q

c

1

= \frac{Q}{2} c_2

2

Q

c

2

=> C_1C

1

= C_2C

2

Initial energy stored on the capacitor C_1C

1

= \frac{ {Q}^{2} }{2C_1}

2C

1

Q

2

Final energy on the capacitor when charge has been halved = \frac{ \frac{Q}{2}^2 }{2C_1}

2C

1

2

Q

2

= \frac{ {Q}^{2} }{4C_1}

4C

1

Q

2

Final energy on the capacitor C_2C

2

= \frac{ \frac{Q}{2}^2 }{2C_2}

2C

2

2

Q

2

= \frac{ \frac{Q}{2}^2 }{2C_1}

2C

1

2

Q

2

( Since, C_1C

1

= C_2C

2

)

Hence, Total final energy = \frac{ {Q}^{2} }{2C_1}

2C

1

Q

2

= Total initial energy.

\boxed { \pink { Hence,\:the\:ratio\:is\:1.} }

Hence,theratiois1.