The circuit in the figure ( http://s19.postimage.org/snuxywi1v/P050figure.png )h

Question

The circuit in the figure (http://s19.postimage.org/snuxywi1v/P050figure.png)has a capacitor, C = 4.50 mF, connected to a V = 6.0-V battery, two switches, and three resistors (R1 = 340 ?, R2 = 220 ?, and R3 = 140 ?. Initially, the capacitor is uncharged and both of the switches are open.

a) Switch S1 is closed. What is the current flowing out of the battery immediately after switch S1 is closed?

b) After 10.0 min, switch S2 is closed. What is the current flowing out of the battery immediately after switch S2 has been closed?

c) What is the current flowing out of the battery 10.0 min after switch S2 has been closed?

d) After another 10.0 min, switch S1 is opened. How long will it take until the current in the R2 = 220-? resistor is below 1.00 mA?

Explanation / Answer

Please ask if you have any doubt.I will help you.

a) I = 6V / (340 + 140) = 12.5 mA

b) The capacitor acts as short initially .

I = 6V / (340 + 140 II 220) = 6V / 425.56

I = 14.1 mA

c) The time constant is = Req C

Req = 340 II 140 + 220 = 319.17

= 319.17 (4.5mF) = 1.44 s.

10 min a very long time .The capactior acts as open circuit.

I = 6V / 480 = 12.5 mA.

d) if S1 is opened , the time constant

1 = 360 4.5 mF = 1.62 s

Voltage across the C initially = 6V 140/480 = 1.75 V

Initial Current when S1 is opened = 1.75 V / 360 = 4.86 mA

i(t) = 4.86 e-t/1.62s mA.

1mA = 4.8 e-t/1.62s mA gives

t = 2.54 s.

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