A capacitor 10 mu F is running in a periodic mode it has acc

A capacitor (10 mu F) is running in a periodic mode it has accumulated total charge of 0.4 mu C. We know the lowest the voltage of the capacitor is 10 V. What is the peak voltage of the capacitor? 10V 10.4 V 10.04V 11V

Solution

Charge on capacitor=0.4uC, capacitance=10uF.

So, voltage=Q/C=0.4/10=0.04V

Given, lowest voltage on capacitor=10V

So peak voltage=10+0.04=10.04V