EELE 250 practice final exam solution summary
1) Vth: solve circuit for voltage between upper right node and bottom node. 24
volts;
Rth: turn "off" independent voltage and current sources and calculate resistance
looking into the port. Rth = 5 ohm || 20 ohm = 4 ohm
2) Negative feedback, so use ideal assumptions. I1 = +0.2mA; I2 = 0; I3 = +0.2mA (= I1); I4 = -0.7 mA
3) Capacitor voltage starts at zero and cannot change instantaneously, so capacitor current must jump from zero to V/R. So Figure c is the correct solution.
4) Initial inductor current = Vs/R1
Inductor current at inf = 0
Inductor current for t>0: (Vs/R1) x exp{ -t / (L/R) }
5) Working from right to left, notice that voltage from top right node to bottom is 5V, then determine currents. Result is Vx = 25 volts.
6) Note current source in upper mesh, so i1 = - 5 amps (direction opposite i1 direction). Then solving lower mesh gives i2 = 2 amps.
7) Efficiency is Pout/Pin = 85%
8) Speed regulation is 5.3%
9) Vth: no current in elements because Vth is "open circuit" voltage, so Vth = 20 at 0 degrees. Zth is jwL + 1/(jwC) = 90 at -90 degrees.