The switch in the circuit has been open for a long time. The resistors in the ci

Question

The switch in the circuit has been open for a long time. The resistors in the circuit each have a value of 60 Ohms, the inductors each have a value of 270 mH, and V = 10 V. (a) What is the total equivalent inductance of the circuit between points A and B? (b) What is the current through R1 the instant after the switch is closed? (c) What is the current through R1 long after the switch has been closed? R1 and R2 are perpendicular, Inductors 1, 2, and 3 are in series; A is between R2 and I2, I1 separates A and B

Explanation / Answer

A. At t=0, I2 = 0 and V(C) = 0, so I3 = E/(R1+R3) = 8/13 A, thus I1 = 8/13 A B. At t=0, I2 = 0 and V(C) = 0, so V(L) = ER3/(R1+R3) = 64/13 V, and LdI2/dt = V(L) ==> dI2/dt = V(L)/L = 1641.0 A/s C. At t=(large number), I3 = 0 and V(L) = 0, so I2 = E/(R1+R2) = 8/11 A, and U = I2^2*L/2 = 7.9339E-4 J D. At t=(large number), I3 = 0 and V(L) = 0, so V(C) = ER2/(R1+R2) = 48/11 V, and U = V(C)^2*C/2 = 8.5686E-5 J E. Immediately after the switch is opened, current I2 continues to flow, but now it flows in the loop comprising L, C, R3 and R2. This current is in the same (downward) direction through R2 as before, and in the upward direction through R3. I2 = 8/11 A, I3 = -8/11A

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