1. What is the charge on a capacitor of 2 micro farads immediately after a switc

Question

1. What is the charge on a capacitor of 2 micro farads immediately after a switch is closed? There is a 9 V battery on the vertical line to the left of the capacitor and a 50 ohm resistor on the vertical line to the right of the capactior. 2. What is the current thru the resistor after the switch is closed? 3. What is the charge on the capacitor when t=50 micro seconds? 4. What is the current thru the resistor when t = 50 micro Coulombs? 5. What is the charge on the capacitor when t = 200 micro seconds? 6. What is the current thru the resistor when t = 200 micro seconds?

Explanation / Answer

when switch is closed

Cherge on Capacitor is q = C*V*[(1-e^(-t/T)]..

here T = R*C = 2*10^-6*50 = 10^-4 S

immediately after switch is closed means t= 0

q = 2*10^-6*9*[1-e^(-0)]

q = 0 C

2) current i = (V/R)*e^(-t/T)

here t = 0 S

i = (9/50) = 0.18 A

3) q = C*V*[(1-e^(-t/T)]..
t/T = 50*10^-6/(10^-4) = 50*10^-2= 0.5
q = 2*10^-6*9*(1-e^(-0.5))
q = 7.082 micro Coulomb

4) i = (V/R)*e^(-t/T)
i = 0.18*e^(-0.5) = 0.109A

5) t/T = (200*10^-6)/10^-4 = 2
q = C*V*[(1-e^(-t/T)].
q = 2*10^-6*9*(1-e^(-2)) = 15.5 micro coulombs

6) i = (V/R)*e^(-t/T)
i = 0.18*e^(-2)
i =0.0243 A

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