The circuit below shows a capacitor C 10 muF two ideal batt

The circuit below shows a capacitor C = 10 muF, two ideal batteries _2= 3.0 V and _1= 1.0 V. two resistors R_2= 0.4 Ohm and R_1= 0.2 Ohm and a switch S. Initially S has been open for a long time. If it is then closed for a long time, by how much does the charge on the capadtor change?

Solution

When Switch S is open ,the charge on the capacitor initally is

Q_initial =CE₂ =10u*3 =30 uC

When Switch S is closed ,the current flowing in the circuit is

I=(E₁-E₂)/(R₁+R₂) =(1-3)/(0.2+0.4)

I=-3.33 A

Voltage drop across R₁

V₁=IR₁ =-3.33*0.2 =-0.6667 Volts

Voltage drop across capacitor

V_c=E₁-V₁ =1+0.6667 =1.6667 Volts

So final Charge on the capacitor

Q_final =CV_c = 10*1.6667 =16.67 uC

Difference in charge is

Q_initial-Q_final = 30-16.67 =13.33 uC