Transient Analysis Questions and Answers

Answer: B

The inductor current at \(t = 0^{+}\) is 1 A.

Before the switch operates the circuit is in steady state, and an ideal inductor carrying direct current behaves as a short circuit. So the whole 20 V appears across the 20 Ω resistor:

\[ i_{L}(0^{-}) = \frac{20}{20} = 1\ \text{A} \]

The current through an inductor cannot change instantaneously, because that would need infinite voltage, so \(i_{L}(0^{+}) = i_{L}(0^{-}) = 1\) A. Closing the switch across the inductor only lets that current circulate through the switch, where it stays in the ideal case.

0 A ignores the continuity of inductor current, and 0.5 A or 2 A do not follow from the circuit values.

Answer: B

The current at \(t = 0\) is 10 A and the final capacitor voltage is 200 V.

At the instant the switch closes, the capacitor already holds 100 V opposing the 200 V source, and the diode is forward biased because the source is the higher. The net voltage drives the current through R:

\[ i(0) = \frac{200 - 100}{10} = 10\ \text{A} \]

The capacitor then charges until its voltage equals the source voltage, 200 V, at which point the current falls to zero; the diode stops it from reversing. 20 A ignores the initial charge on the capacitor, and 150 V is only the average of the two voltages, not where the charging stops.

Answer: A

The time constant is \(\tau = \frac{L}{R}\).

When a dc voltage is switched onto an R-L series circuit, the current rises as \(i = I\left(1 - e^{-Rt/L}\right)\). The time constant is the time in which the current would reach its final value if it kept its initial rate of rise; in practice it is the time to reach 63.2 per cent of the final value. Henry divided by ohm gives seconds.

R/L is the reciprocal, with units of 1/second. RL has units of ohm-henry, not time, and 1/RL is no better. The R-C circuit's time constant is RC, which may be the source of the RL trap.