The switch in the figure (Figure 1) has been in position a for a long time. It i

Question

The switch in the figure (Figure 1) has been in position a for a long time. It is changed to position b at t=0s

Part A

What is the charge Q on the capacitor immediately after the switch is closed?

Part B

What the current I through the resistor immediately after the switch is closed?

Part C

What is the charge Q on the capacitor at t=50s?

Part D

What is the current I through the resistor at t=50s?

Part E

What is the charge Q on the capacitor at t=200s?

Part F

What is the current I through the resistor at t=200s?

Explanation / Answer

Solution-

a)The charge on the capacitor is given by
q0 = CV = (4*10^-6F)(9V) = 36*10^-6C
b)he current through the resistor is i0 = q/RC
   = 36*10^-6 / (25)(4*10^-6)
   = 0.36A
c) The charge on the capacitor after t = 50*10^-6s
q = q0 e^(-t/RC)
   = 36*10^-6 e^( -50*10^-6 /(25)(4*10^-6))
   = 21.8*10^-6C
d) The current
i = i0 e^(-t/RC)
   = 0.36e^( -50*10^-6 /(25)(4*10^-6))
   = 0.22 A
e) The charge on the capacitor at t = 200 µs
      q = q0 e^(-t/RC)
   = 36*10^-6 e^( -200*10^-6 /(25)(4*10^-6))
   = 4.87*10^-6C
f) The current after t = 200µs
   i = i0 e^(-t/RC)
   = 0.36e^( -200*10^-6 /(25)(4*10^-6))
   = 0.048 A

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