Consider the simple RC circuit shown in Figure 7. The battery emf is epsilon = 1

Question

Consider the simple RC circuit shown in Figure 7. The battery emf is epsilon = 15.0 V, the capacitance is C = 0.60 mu F & the resistance is R = 30 k Ohm. The switch S is closed, a current I begins to flow & a charge Q begins to accumulate on the capacitor plates. Calculate: The time constant tau of this circuit. The maximum charge Q_m that the capacitor could acquire. The time it takes for the capacitor charge Q to reach a value Q = 0.9 Q_m. The maximum value of the current I_m that will flow in the circuit. The time it takes for the current I to reach a value I = 0.7 I_m.

Explanation / Answer

a)

Time Constant

T=RC =30000*(0.6*10-6)

T=0.018 s or 18 ms

b)

Maximum Charge

Qmax=CV =(0.6*10-6)*15

Qmax=9*10-6 C or 9 uC

c)

In a RC circuit charge as a funciton of time is given by

Q=Qmax[1-e-t/T]

0.9Qmax=Qmax[1-e-t/0.018]

ln(0.1)=-t/0.018

t=0.04145 s or 41.45 ms

d)

Maximum Current

Imax=E/R =15/30000

Imax=5*10-4 A or 0.5 mA

e)

In a RC circuit current as a funciton of time is given by

I=Imaxe-t/T

0.7Imax=Imaxe-t/0.018

-t/0.018 =Ln(0.7)

T=0.00642 s or 6.42 ms

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