Question

The circuit of the figure shows a capacitor, two ideal batteries, two resistors, and a switch S. Initially S has been open for a long time. If it is then closed for a long time, what is the change in the charge on the capacitor? Assume C = 43 μF, 1 = 1.7 V, 2 = 6.0 V, R1 = 0.41 Ω, and R2 = 0.59 Ω.

Answer 1

When the switch is open, the capacitor is charged by battery &z only through the resistance R. The initial charge is, 9: = CEZ = ( 43 4F)(6 V) = (43x10F)(6V) = 258x106c When the switch is closed, the current i is, i(R+R)= {2- & is Ez- & (R+R) 6 V-1.7 V i= 0.41 2+0.59 22 = 4.3 A The potential difference across the capacitor is, V = {2 - IR = 6 V-(4.3 A)(0.592) = 3.643 V The final charge is, 9; = CV = (43 4F)(3.643 V) = (43x10^ F)(3.643 V) = 148.909 x10-60 Calculate the change in charge as follows: Aq = 9;-9 = 148.909x10-C-258x10-6 C =-109.091x106c -110x10c (rounding off to two significant figures)