Question

Consider a RC circuit with Vi 20 V, R, = 100 and R2 = 200 . The capacitors are C, = 12 uF and C2 = 40 uF. The switch in the Figure 6(b) has been closed at position a for a long time. b а R Figure 6(b) What is the amount of electrical energy stored in the capacitor? (i) ANS: The switch is now flipped to position b and remains in position b for a long (ii) time. What fraction of the initial charge is in C after a long time?

Answer 1

i) after a lont time the capacitor acts as open circuit.

so, Req = R1 + R2

= 100 + 200

= 300 ohms

current through the battery, I = V/Req

= 20/300

= 0.0667 A

potential drop across C1 = potential drop across R2

= I*R2

= 0.0667*200

= 13.3 V

so, charge on C1, Q1 = C1*13.3

= 12*13.3

= 160 micro C

electrical energy stored in C1, U1 = (1/2)*C1*V1^2

= (1/2)*12*10^-6*13.3^2

= 1.06*10^-3 J <<<<<<<-----------------Answer

ii) when switch is closed to b position.

Ceq = C1 + C2

= 12 + 40

= 52 micro F

potential drop across the two capacitors, V = Q1/ceq

= 160/52

= 3.08 V

New charge on C1, Q1' = C1*V

= 12*3.08

= 37 micro C

Q1'/Q1 = 37/160

Q1' = 0.231*Q1 <<<<<<<-----------------Answer

new charge on C1 = -/231 times initial charge on C1