Work done by 1 V battery when switch S is closed in - J is-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Capacitance of Parallel Plate Capacitor

Work done by ...

Question

Work done by 1 V battery when switch S is closed in $μ J$ is .

Open in App

Solution

Initial charge on left plate of 3 $μ$ F is -8 $μ C$ .
Charge on left plate of $3 μ$ F after closing switch is $- 3 μ C$ .
Therefore, charge flown through 1V after closing switch is $- 5 μ C$ . So, work done by the battery is $W = Q \times V = - 5 μ \times 1 = - 5 μ J$

Suggest Corrections

Similar questions

Work done by 1 V battery when switch S is closed in

μ J

is .

A 4 $μ$ F capacitor is given 20 $μ$ C charge and is connected with an uncharged capacitor of capacitance 2 $μ$ F as shown in figure. When switch S is closed –

In the circuit shown is figure, switch is closed then find the work done by battery till next steady state is achieved.

Q. Consider the situation shown in figure, the switch

S

open for a long time and then closed. Then work done by the battery will be:

Q. Consider the situation shown in figure (31-E23). The switch S is open for a long time and then closed. (a) Find the charge flown through the battery when the switch S is closed. (b) Find the work done by the battery.

Join BYJU'S Learning Program

Work done by 1 V battery when switch S is closed in μJis .

Initial charge on left plate of 3μF is -8μC. Charge on left plate of 3μF after closing switch is −3μC. Therefore, charge flown through 1V after closing switch is −5μC. So, work done by the battery is W=Q×V=−5μ×1=−5μJ

Work done by 1 V battery when switch S is closed in $μ J$ is .

Initial charge on left plate of 3 $μ$ F is -8 $μ C$ .
Charge on left plate of $3 μ$ F after closing switch is $- 3 μ C$ .
Therefore, charge flown through 1V after closing switch is $- 5 μ C$ . So, work done by the battery is $W = Q \times V = - 5 μ \times 1 = - 5 μ J$