The circuit shown above contains two switches; both areinitially open and the ca

Question

The circuit shown above contains two switches; both areinitially open and the capacitor is uncharged. The battery has zerointernal resistance. The parameters of the circuit elementsare:

E = 10 V
R1 = 4 × 104W
R2 = 3 × 104W
R3 = 20 × 104W
C = 5 µF

At t = 0 switch S1 is closed (S2remains open).

(a) Calculate the time constant that applies tocharging of the capacitor.

tc = s    

(b) Calculate the charge on the capacitor after a very long time(many time constants).

Qmax = C    

(c) Calculate the charge on the capacitor at the following tmes:t = 0.08 s, 0.25 s, and 0.6 s.

Q(0.08 s) = C   

Q(0.25 s) = C   

Q(0.6 s) = C   

(d) Calculate the current flowing into the positive plate of thecapacitor at the following times: t = 0.08 s, 0.25 s, and0.6 s.

I(0.08 s) = A   

I(0.25 s) = A   

I(0.6 s) = A   

Next switch S1 is opened and following that, switchS2 is closed. Let Dtrepresent the time interval after S2 is closed.

(e) Calculate the time constant that applies todischarging of the capacitor.

td = s    

(f) Calculate the charge on the capacitor at the followingtimes: Dt = 0.46 s, 1.4 s, and2.5 s, where Dt denotes thetime since S2 was closed.

Q(0.46 s) = C   

Q(1.4 s) = C   

Q(2.5 s) = C

Explanation / Answer

. 01.   Part a: 02.      c = (R1 +R2)*(C) = (40000 + 30000)*(5.0E-06 Farads) =0.350 seconds 03.   Part b: 04.      Q =C*Vc 05.      Vc =(Vsource)*[1 - exp(-t/c)] = (10volts)*{1 - exp[(-t)/(0.350 seconds)]} 06.      Q = (5.0E-06 Farads)*(10volts) = 50.0E-06Coulombs 07.   Part c: 08.      Q(t = 0.08 seconds) =(5.0E-06 Farads)*(10 volts)*{1 - exp[(-0.08 s)/(0.350 s)]} =10.22E-06Coulombs 09.      Q(t = 0.25 seconds) =(5.0E-06 Farads)*(10 volts)*{1 - exp[(-0.25 s)/(0.350 s)]} =25.52E-06 Coulombs 10.      Q(t = 0.60 seconds) =(5.0E-06 Farads)*(10 volts)*{1 - exp[(-0.60 s)/(0.350 s)]} =41.00E-06 Coulombs 11.   Part d: 12.      VR =(V)*[exp(-t/)] = (10 volts)*{exp[(-t)/(0.350s)]} 13.      I = (VR)/(R) =[(10 volts)/(70000)]*{exp[(-t)/(0.350s)]} 14.      I(t = 0.08 seconds) =[(10 volts)/(70000)]*{exp[(-0.08s)/(0.350s)]} = 113.7E-06 Ampere 15.      I(t = 0.25 seconds) =[(10 volts)/(70000)]*{exp[(-0.25s)/(0.350s)]} = 69.93E-06 Ampere 16.      I(t = 0.60 seconds) =[(10 volts)/(70000)]*{exp[(-0.60s)/(0.350s)]} = 25.73E-06 Ampere 17.   Part e: 18.      d = (R3)*(C)= (200000)*(5.0E-06 Farads) = 1.00 seconds 19.   Part f: 20.      Vd =(Vc_max)*[exp(-t/d)] = (10volts)*{exp[(-t)/(1.00 seconds)]} 21.      Q(t = 0.46s) =C*Vd = (5.0E-06 Farads)*(10 volts)*{exp[(-0.46s)/(1.00seconds)]} = 31.56E-06Coulombs 22.      Q(t = 1.40s)  =(5.0E-06 Farads)*(10 volts)*{exp[(-1.40s)/(1.00 seconds)]} =12.33E-06Coulombs 23.      Q(t = 2.50s)  =(5.0E-06 Farads)*(10 volts)*{exp[(-2.50s)/(1.00 seconds)]} =4.104E-06Coulombs .

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