Consider the following Series RC circuit Assume that there i

Consider the following Series RC circuit. Assume that there is no initial charge on the capacitor before the switch is moved to the upward (charging) position. Also assume that t=0 at the instant the switch first makes contact. Answer the following questions. Be sure to show the formulas you use, any essential algebraic or calculus steps, and where each numerical value is placed into the formulas. Also, remember to give magnitudes and directions for the currents (draw a picture with an arrow showing the current direction) and magnitudes and polarities for the voltages (draw a picture with \"+\" and \"-\"signs showing the polarity)! Finally, remember to include units in your final results! What is the voltage across the resistor V_a immediately after closing the switch? What is the voltage across the capacitor V_c immediately after closing the switch? At what time will the energy stored in the capacitor be 0.01 Joules? What is the voltage across the resistor V_g at 4.00s? What is the voltage across capacitor V_c at 4.00s?. How do the Emf, VR and V_c relate to each other at 4.00s and why? What power is being supplied by the Emf at 4.00s? What power is being dissipated by the resistor at 4.00s? At what rate is energy being stored in the capacitor at 4.00s? How does the power being supplied by the Emf relate to the rate at which energy is being stored in the capacitor and the power being \"dissipated\'\' by the resistor at 4.00s?

Solution

a) PD across capacitor can not change instantly.

hence just after closing switch,

PD across capacitor will be zero.

henceVC = 0

Vc + Vr = e

Vr = 20 volt

b) Vc = 0 (as discussed in a. )

c) E = C V^2 / 2

   0.01 = (200 x 10^-6) V^2 / 2

V = 10 V

Vc = E [ 1 - e^(-t/RC) ]

10 = 20 [1 - e^(- t / (200 x 15 x 10^3 x 10^-6)) ]

- t / 3 = ln(0.5)

t = 2.08 sec

d) Vr = E e^(-t/RC)

Vr = 20 e^(-4 / (3)) = 5.27 volt

e) Vr + Vc = e

5.27 + Vc = 20

Vc = 14.73 volt

f) Vr + Vc - R = 0   ( applying KVL in the loop )

g) P = E I

and I = Vr / R

P = (20 ) ( 5.27 / 15 x 10^3) = 7.03 x 10^-3 W = 7.03 mW

h) P = Vr^2 / R = 5.27^2 / (15 x 10^3) = 1.85 x 10^-3 W Or 1.85 mW

i) Pc = 7.03 - 1.85 = 5.18 mW