Sinusoidal Steady State Analysis Questions and Answers

Answer: B

The real power is \(P = V_{rms} I_{rms}\cos\phi\). The supply voltage has a peak of 100 V, so

\[ V_{rms} = \frac{100}{\sqrt{2}} = 70.7\ \text{V} \]

\[ \begin{aligned} I_{rms} = \frac{P}{V_{rms}\cos 30^\circ} \\ &= \frac{400}{70.7 \times 0.866} = 6.53\ \text{A} \end{aligned} \]

The current lags the voltage by \(30^\circ\), so its angle is \(-30^\circ\) relative to the voltage. The key gives 6.53 A at \(-30^\circ\), option B; strictly the phase with respect to the reference would be \(10^\circ - 30^\circ = -20^\circ\), but none of the options says that, so the angle is read as the lag. The values of 65.3 A come from forgetting the rms conversion by a factor of ten.

Answer: C

For a single phase load the real power is \(P = VI\cos\phi\), so the line current is

\[ \begin{aligned} I = \frac{P}{V\cos\phi} \\ &= \frac{50 \times 10^3}{230 \times 0.5} \\ &= \frac{50000}{115} = 434.8\ \text{A} \end{aligned} \]

This is the value of 434.7 A in the options, option C. A current of 43.47 A or 4.34 A would be a slip of one or two powers of ten, and 0.4347 A would come from taking the power as 50 W. The current lags the voltage by \(\cos^{-1}0.5 = 60^\circ\).

Answer: C

In a star connection the line voltage is \(\sqrt{3}\) times the phase voltage.

Each line voltage is the phasor difference of two phase voltages that are 120° apart. The difference of two equal phasors at 120° has a magnitude of \(2V_{ph}\cos 30^{\circ} = \sqrt{3}V_{ph}\), and it leads the corresponding phase voltage by 30°. The line current, on the other hand, equals the phase current.

\(1/\sqrt{3}\) is the ratio the other way round (phase to line voltage), and \(\sqrt{2}\) relates peak and rms values, not star line and phase voltages.

Answer: C

A low power factor means the circuit will draw more reactive power.

Power factor is the ratio of active power to apparent power. For a given active power, a low power factor means a larger apparent power, the difference being reactive power that flows back and forth between the source and the load's inductance. The circuit therefore draws more reactive power and a larger current.

Because the current is larger, the line voltage drop and losses increase, so 'less line current' and 'less voltage drop' are the opposite of what happens. The active power is set by the load and does not grow because the power factor is poor.

Answer: D

The active power is 195 kW.

Active power is apparent power multiplied by the power factor.

\[ P = 300 \times 0.65 = 195\ \text{kW} \]

228 kW is the reactive power, \(300 \sin\phi = 300 \times 0.76\), given in kVAR rather than kW, and 285 and 185 kW correspond to no standard calculation from these figures.

Answer: A

The current is \(jV\omega C\).

The impedance of a capacitor is \(Z_{C} = \frac{1}{j\omega C} = -\frac{j}{\omega C}\). Taking the voltage V as the reference phasor:

\[ I = \frac{V}{1/j\omega C} = j\omega C V \]

The factor j shows the current leading the voltage by 90°, as it does in a capacitor, and its magnitude is \(\omega C V\). \(-jV\omega C\) would lag (an inductor's behaviour), \(jV/\omega C\) has the wrong magnitude, and \(V/j\omega C\) multiplies by the impedance instead of dividing.

Answer: C

The form factor is rms value / average value.

Form factor describes the shape of a waveform as the ratio of its rms value to its average (half-cycle mean) value. For a sine wave it is 0.707V_m / 0.637V_m = 1.11, which is why average-responding meters are scaled by 1.11 to read rms.

Maximum value divided by rms value is the peak (crest) factor, 1.414 for a sine wave. The ratios with average or rms value over maximum value, and average over rms, are not the form factor.

Answer: D

The current drawn is 5.77 A.

In a star connection the line current equals the phase current, and each phase sees the phase voltage, the line voltage divided by \(\sqrt{3}\).

\[ \begin{aligned} V_{ph} &= \frac{400}{\sqrt{3}} \approx 230.9\ \text{V} \\ |Z| &= \sqrt{32^{2} + 24^{2}} = 40\ \Omega \\ I &= \frac{230.9}{40} \approx 5.77\ \text{A} \end{aligned} \]

10 A uses the full line voltage across one phase, which is correct only for a delta load's phase current, and 17.32 A is the line current that delta load would draw. 20.50 A does not come from these values.

Answer: A

The supply frequency is about 63.6 Hz.

The reactance of a capacitor is \(X_{C} = 1/(2\pi f C)\), so the frequency follows directly; the 100 V supply voltage is not needed.

\[ \begin{aligned} f &= \frac{1}{2\pi C X_{C}} \\ &= \frac{1}{2\pi \times 50 \times 10^{-6} \times 50} \\ &= \frac{1}{0.0157} = 63.66\ \text{Hz} \end{aligned} \]

At 50 Hz the reactance of 50 µF would be about 63.7 Ω, not 50 Ω. A capacitor at zero frequency (dc) has infinite reactance.

Answer: C

The frequency is 25 Hz.

An alternating voltage written as \(v = V_{m}\sin(\omega t)\) has angular frequency \(\omega = 2\pi f\). Here \(\omega = 157\) rad/s.

\[ \begin{aligned} f &= \frac{\omega}{2\pi} = \frac{157}{2 \times 3.14} \\ &= \frac{157}{6.28} = 25\ \text{Hz} \end{aligned} \]

50 Hz would need \(\omega = 314\) rad/s, and 100 Hz 628 rad/s. 75 Hz does not correspond to 157 rad/s at all.