Home
Class 12
PHYSICS
In the given figure switch is open initi...

In the given figure switch is open initially and capacitor `C_2` is unchanged

Energy in C1 is

A

`CV^(2)`

B

`(1)/(2)CV^(2)`

C

`(1)/(4)CV^(2)`

D

`(1)/(8)CV^(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
Promotional Banner

Similar Questions

Explore conceptually related problems

In the given figure switch is open initially and capacitor C_2 is unchanged Charge on C2 just after closing the switch is

In the given figure switch is open initially and capacitor C_2 is unchanged Potential difference on C1just after closing the switch is

In the circuit shown in the figure, the switch S is initially open and the capacitor is initially uncharged. I_(1),I_(2) and I_(3) represent the current in the resistance 2Omega, 4Omega and 8Omega respectively.

In the given LC circuit,if initially capacitor C has charge Q on it and 2C has charge 2Q. The polar ar as shown in figure. Then after closing switch S and t=0

In the connection shown in the figure, initially the switch K is open and the capacitor is uncharged. Then the switch is closed, and the capacitor is charged up to the steady state and the switch is opened again. Determine the values indicated by the values indicated by the ammeter just after closing the switch [Given V_0 = 30 V, R_1 = 10k Omega, R_2 = 5kOmega ] . A long time after switch was closed.

In the connection shown in the figure, initially the switch K is open and the capacitor is uncharged. Then the switch is closed, and the capacitor is charged up to the steady state and the switch is opened again. Determine the values indicated by the values indicated by the ammeter. [Given V_0 = 30 V, R_1 = 10k Omega, R_2 = 5kOmega ] . A long time after the switch was closed.

In the connection shown in the figure, initially the switch K is open and the capacitor is uncharged. Then the switch is closed, and the capacitor is charged up to the steady state and the switch is opened again. Determine the values indicated by the values indicated by the ammeter. [Given V_0 = 30 V, R_1 = 10k Omega, R_2 = 5kOmega ] . A long time after the switch was closed.

In the given circuit switch K is open. The charge on the capacitor is C is steady-state is q_1 Now the key is closed and steady-state charge on C is q_2 The ratio of charges q_1//q_2 is -

Two capacitors C_1 = C and C_2 = 3C are connected as shown in figure, Initially, key K is open and capacitor C_1 holds charge Q. After closing the key K, the charge on each capacitor at steady state will be

In the given circuit, in steady state we may say: (capacitors are initially unchanged)