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A series RC circuit has a time constant of 1.0 s. The battery has a voltage of 50 V and the maximum curent just after closing the switch is 500 mA. The capacitor is initially uncharged.

(a.) What is the charge on the capacitor 2.0 s after the switch is closed?

(b.) What is the current running through the resistor 2.0 s after the switch is closed.
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as V =iR

R = V/i = 50/0.5 = 100 ohms

now apply Time constant T = RC

so C = 1/100 = 10 mF

so charge Q = CV

Q = 10mF * 50 = 0.5 C

q = Q(1-e^-t/Rc)

q = 0.5*(1-e^-2/1)

q = 0.432 C

--------------------------------------------

apply I= I0 e^-t./RC

I = 500mA*e^(-2)

I = 67.67 mA
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