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! 1. 0 R=15 k E = 20V | + C=200 F a. What is the voltage across the resistor V immediately after closing the switch? b. What is the voltage across the capacitor Ve immediately after closing the switch? c. At what time will the energy stored in the capacitor be 0.01 Joules? d. What is the voltage across the resistor VR at 4.00s? e. What is the voltage across capacitor Ve at 4.00s? f. How do the Emf, VR, and Vc relate to each other at 4.00s and why? g. What power is being supplied by the Emf at 4.00s? h. What power is being dissipated by the resistor at 4.00s? i. At what rate is energy being stored in the capacitor at 4.00s? j. 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

Solution

charging of capicator is already answered now move to discharge of capacitor.

k) after 6 second of charging to stored charge is

Q=Qmax[1-Exp(-t/(CR))]

Qmax=CVapplied = 200 x 10-6 x20 =0.004 columb

Q =0.004[ 1- exp(-6/(200x 10-6 x15x103)

Q=0.003458 Columb

l) equation of capicator discharge

Q=Qmax exp(-t/RC)

differenciate it

I=Qmax/(RC)exp(-t/RC)

at t=0

I=Qmax/(RC) =0.00115 ampere

Vr =IR =17.29 Volt

m)Q=CV

V =0.003458/(200 x 10^-3) =17.29 volt

n) E=Q²/(2C)

0.01 =Q2/(2x200 x 10^-6)

Q=0.002 column

Q=Qmax exp(-t/RC)

0.002=0.003458 exp(-t/RC)

t=1.64 second

o) I=Qmax/(RC)exp(-t/RC)

I =0.000304 ampere

V=IR =4.558

p) From diagram VC=Vr =4.558

r/s) power supplied by capicator =power dissipated by ressitor = Ix I XR =0.001385

t/u )current is flow in upward direction in resistor and top plate of capictor is postively charged

v) I=Qmax/(RC)exp(-t/RC)

I=0.25/1000

t=4.58 s