Sinusoidal Steady State Analysis Questions and Answers

Answer: B

The average power is 12 W.

The source has a peak value of 12 V, and average power in a resistor depends on the rms value, \(P = V_{rms}^{2}/R\). For a pure resistor the frequency (377 rad/s, that is 60 Hz) does not matter.

\[ \begin{aligned} V_{rms} &= \frac{12}{\sqrt{2}} \\ P &= \frac{V_{rms}^{2}}{R} = \frac{144/2}{6} = 12\ \text{W} \end{aligned} \]

Using the peak value in place of the rms value gives 144/6 = 24 W, which is twice the true average. 4 W, 14 W and 16 W do not follow from the circuit.

Answer: B

A power factor of 1 indicates a purely resistive circuit.

Power factor is cos φ, where φ is the phase angle between voltage and current. It equals 1 when φ = 0, that is, when the voltage and current are in phase and the circuit as a whole behaves as a pure resistance. This happens in a circuit of resistors alone, and also in an RLC circuit at resonance, where the inductive and capacitive reactances cancel, which is why the answer speaks of the circuit rather than a single element.

A purely reactive element or circuit has φ = 90° and a power factor of 0, so those options describe the opposite case.

Answer: C

The load resistor is R.

The output is taken across C in parallel with \(R_{L}\), fed through the series resistor R, so the circuit is a voltage divider.

\[ \begin{aligned} Z_{p} &= \frac{R_{L}}{1 + sCR_{L}} \\ \frac{V_{o}}{V_{i}} &= \frac{Z_{p}}{R + Z_{p}} \\ &= \frac{R_{L}}{R(1 + sCR_{L}) + R_{L}} \\ &= \frac{1}{1 + R/R_{L} + sCR} \end{aligned} \]

For this to equal \(\frac{1}{2 + sCR}\), the constant terms must match: \(1 + R/R_{L} = 2\), so \(R_{L} = R\).

With \(R_{L} = R/2\) the denominator would start with 3, with \(R_{L} = 2R\) it would start with 1.5, and with \(R/4\) it would start with 5.

Answer: B

The poles are at s = -2 and s = ∞.

Looking into the input, the 2 H inductor is in series with the 1 Ω shunt resistor in parallel with the branch of the 1 H inductor and the second 1 Ω resistor:

\[ \begin{aligned} Z(s) &= 2s + \frac{1 \cdot (s + 1)}{1 + s + 1} \\ &= 2s + \frac{s + 1}{s + 2} \\ &= \frac{2s^{2} + 5s + 1}{s + 2} \end{aligned} \]

The denominator gives a finite pole at s = -2. The numerator is of higher degree than the denominator, so Z(s) also grows without limit as s goes to infinity, which is a pole at infinity. Options without ∞ miss that pole, and +2 has the wrong sign: an RL network has its poles in the left half-plane.