Question

a capacitor of 10 micro farad capacity is connected to 12V Battery and a resistance. Calculate work done by battery to increase the potential of capacitor from 5V to 5.1V​

Answer 1

Answer:

I don't know how to calculate it

Answer 2

The work done by the capacitor is w= 12μJ

• The voltage across the capacitor is to be increased from 5V to 5.1V.
• Let w be the work done
• The change in potential is 5.1 - 5 = 0.1 V
• Equation for work done is,  W=q ΔV
• Charge in the capacitor is calculated using the formula,  q=C V

                                         q=C V

                                           = 10*$10^{-6}$ *12

                                           =1.2*$10^{-4}$ c

• work done              W=1.2* $10^{-4}$ *0.1
•                                    = 1.2*$10^{-5}$ J
•                                    =12 μJ