1)The switch has been open for a long time when at time t = 0, the switch is clo

Question

1)The switch has been open for a long time when at time t = 0, the switch is closed. What is I4(0), the magnitude of the current through the resistor R4 just after the switch is closed?in Amps

2)What is I4(?), the magnitude of the current through the resistor R4 after the switch has been closed for a very long time? in Amps

3)What is IL(?), the magnitude of the current through the inductor after the switch has been closed for a very long time? in Amps

4)After the switch has been closed for a very long time, it is then opened. What is I3(topen), the current through the resistor R3 at a time topen = 3.3 ms after the switch was opened? The positive direction for the current is indicated in the figure.in Amps

5)What is VL,max(closed), the magnitude of the maximum voltage across the inductor during the time when the switch is closed?in Volts

6)What is VL,max(open), the magnitude of the maximum voltage across the inductor during the time when the switch is open?in Volts

Explanation / Answer

1. Just after closing the switch the inductor will act as an open circuit, that means no current will pass through the inductor just after closing the switch. So current will pass only through the branch consisting R3. Then equivalent resistance of the circuit is Req = R1+R3+R4 = 36+107+94 ohm or 237 ohm. So current through R4 or I4 is 12/237 = .0506 A

2. After the switch has been closed for a very long time, the circuit will operate in its steady state. In steady state the inductor will act like a short circuit, that means just like a wire. Then equivalent resistance of the circuit Req = R1+(R2*R3)/(R2+R3) + R4 = 36 + (36*107)/(36+107) + 94 = 156.93 or 157 ohm. So current through R4 or I4 is 12/157 = .076 A.

3. From (2) the total current flowing in the circuit is .076 A and this will get deistributed in the branches consisting R2,R3. So current in through the inductor( or R2) = total current*(value of the other resistance/sum of the resistance)= .076 * (107/(107+36))=.056 A.

4. Inductor doesn't allow to current to change abruptly, so when the switch is opened again, the current flowing through the inductor will flow through R3 just after opening the switch. So the current through R3 just after opening the switch is .056 A.

Time constant of the circuit after opening the switch = L/(R3+R2) = 2.07 * 10^(-3) s

Therfore current after 3.3 ms after opening the switch = (current at t=0)* e ^(-t/time constant),( t = 3.3 * 10^-3 s) == .056 * e ^((-3.3 * 10^-3 )/(2.07 * 10 ^ -3))=.011 A

5. As I said in (1) the inductor will act like an open circuit just after closing the switch, so voltage across it will be 0 .

6. Maximum voltage across the inductor will be just after opening and is equal to .056 * (107+36) = 8.008 V.

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