Single Phase AC Circuits Questions and Answers

Answer: A

The three lamps are connected in parallel across the same AC supply, and in a parallel connection the voltage across every branch is the same as the supply voltage. The voltmeter reading across Lamp1 is 100 V, so the same voltmeter connected across Lamp3 also reads 100 V. The value 50 V would be correct for two equal lamps in series, 300 V would be the sum for three lamps in series and 33.33 V would be a third of 100 V, and none of those is the case. The answer is 100 V, option (A).

Answer: C

To detect a faulty ceramic disc or paper capacitor the preferred test is the charging-holding test: the capacitor is charged from a source and then disconnected, and a good capacitor holds its charge for a time, while a leaky or shorted one loses it quickly. An open circuit test or short circuit test only shows a gross open or short, and the capacitor action-normal charging test is used with a meter on large electrolytic types. The answer is Charging-holding test on capacitors, option (C).

Answer: D

The inductive reactance of a coil is 2 pi f L, where f is the frequency in hertz and L is the inductance in henry.

\[ X_{L} = 2\pi f L = 2\pi \times 1000 \times 0.05 \]

\[ X_{L} = 314.15\ \Omega \]

The reactance is therefore 314.15 ohms, option (D). The inductance of 50 mH is 0.05 H, and the value 3.14 ohms comes from using 0.0005 H, while 50 and 101 ohms do not follow from the formula.

Answer: A

For capacitors used in space electronics the key gives tantalum pentoxide as the dielectric: tantalum capacitors have a very high capacitance for their size and weight, with a stable and long-lasting oxide film, which is needed in space. Impregnated paper, metallised polypropylene and glass are used in other capacitors but are bulkier or less suited to the weight and reliability needs. The answer is tantalum pentoxide, option (A), and the question is a recall item.

Answer: B

The power consumed by an inductor with some resistance is only the power lost in its resistance, since the pure inductance stores energy in its field and returns it to the source.

\[ P = I^{2} R = 13^{2} \times 10 \]

\[ P = 169 \times 10 = 1690\ \mathrm{W} \]

So the power consumed is 1690 W, option (B). The value 1960 W comes from swapping the digits, and the kilowatt values are a thousand times too large.

Answer: B

The impedance of a coil is the square root of the sum of the squares of the resistance and the reactance, since the two are at right angles in the impedance triangle.

\[ Z = \sqrt{R^{2} + X_{L}^{2}} \]

\[ Z = \sqrt{4^{2} + 50^{2}} = \sqrt{2516} \]

\[ Z \approx 50.16\ \Omega \]

So the impedance is about 50.15 ohms, option (B), which is only a little more than the reactance because the resistance is small. Values of 110.12, 25 and 33 ohms do not follow from the formula.

Answer: A

The circuit has a capacitor and a 50 ohm resistor in series on a 12 V supply, with a current of 0.16 A. The voltage across the resistor is 0.16 x 50 = 8 V. In a series RC circuit the supply voltage is the square root of the sum of the squares of the two voltages, so

\[ V_{C} = \sqrt{12^{2} - 8^{2}} = \sqrt{80} \]

\[ V_{C} \approx 8.9\ \mathrm{V} \]

So the voltage across the capacitor is 8.9 V, option (A). The values 18.3 V and 19.2 V are larger than the supply, which cannot be, and 12 V would be the supply voltage itself.

Answer: C

The reactive power is 300 VAR. The apparent, active and reactive powers form a right-angled triangle in which the apparent power is the hypotenuse.

\[ Q = \sqrt{S^2 - P^2} \]

\[ = \sqrt{500^2 - 400^2} = 300\ \text{VAR} \]

The value 100 VAR is the difference of the apparent and active powers (500 - 400), which is not how they combine, and 200 VAR and 400 VAR are not the third side of the 3-4-5 triangle. So the third option is the answer.

Answer: B

The power factor is 0.707. When the active power P and the reactive power Q are equal, the phase angle between voltage and current has tan φ = Q/P = 1, so φ = 45°.

\[ \cos\phi = \cos 45^\circ = \frac{1}{\sqrt{2}} = 0.707 \]

A power factor of 0.5 would belong to an angle of 60° (Q greater than P), 0.6 and 0.8 are the values for a 3-4-5 triangle and would have unequal P and Q. So the second option is correct.

Answer: C

The average power is 32 W. The impedance is (4 - j3) Ω, so the resistance is 4 Ω, and the current has a peak value of 4 A. Only the resistive part takes average power, which is the half of the peak current squared times the resistance.

\[ P = \tfrac{1}{2} I_m^2 R \]

\[ = \tfrac{1}{2} \times 4^2 \times 4 = 32\ \text{W} \]

The value 64 W would result from leaving out the factor of one half, and 16 W is the value of the half applied twice, and 48 W does not satisfy the formula. So the third option is correct.