Exercise practice: Alternating Current

4.1 RMS and power numericals

A 100 W resistor on a 220 V, 50 Hz supply carries an rms current of 100/220 = 0.455 A, which is the value the power formula uses; the peak current is 0.643 A. Over a full cycle the net power consumed is the same 100 W, because the resistor dissipates energy continuously, though the instantaneous power pulses between zero and twice its average.

If the peak voltage of a supply is 300 V, the rms voltage is 300/root 2 = 212 V. If the rms current is 10 A, the peak current is 14.1 A. These conversions are the first step in most exercise numericals.

4.2 Reactance and impedance numericals

A 44 mH inductor on 220 V, 50 Hz: X_L = 314 x 0.044 = 13.8 ohm, so the rms current is 220/13.8 = 15.9 A. A 60 microfarad capacitor on 110 V, 60 Hz: X_C = 1/(377 x 60 x 10^-6) = 44.2 ohm, so the current is 110/44.2 = 2.5 A. In both cases the average power over a full cycle is zero, because the current is a quarter cycle out of phase with the voltage, so the answer to the exercise is zero power with the explanation that energy is stored and returned.

For a series circuit with R = 40 ohm, X_L = 60 ohm and X_C = 30 ohm, the impedance is the square root of 1600 + 900 = 50 ohm, and the phase angle has tan phi = 30/40, so phi is about 37 degrees with the current lagging. If the supply is 220 V rms, the current is 4.4 A and the average power is I squared R = 774 W, which also equals V I cos phi. The same circuit off resonance shows why the power factor matters: the reactive part of the voltage carries no useful power, so the supply must provide a current larger than the resistor alone would need.

4.3 Resonance numericals

A series circuit with L = 27 mH and C = 30 microfarad has a resonant angular frequency of 1 divided by the square root of L C. The product is 27 x 10^-3 x 30 x 10^-6 = 8.1 x 10^-7, whose square root is 9.0 x 10^-4, so omega_0 = 1.11 x 10^3 rad per second and the ordinary frequency is 177 Hz.

At resonance the impedance is R, so the current is maximum and equals V_rms/R. If R = 10 ohm and the supply is 100 V rms, the resonant current is 10 A and the power is 1000 W. Off resonance the current falls, and the width of the peak is set by R: doubling R halves the quality factor and makes the curve flatter.

For a circuit with R = 5 ohm, L = 20 mH and C = 50 microfarad, the resonant frequency is 1 divided by the square root of 10^-6, which is 1000 rad per second. The quality factor is omega_0 L / R = 1000 x 0.02 / 5 = 4.

4.4 Transformer numericals

A step-up transformer has 100 turns in the primary and 1000 in the secondary. With 220 V across the primary, the secondary voltage is 220 x 10 = 2200 V. If the secondary current is 0.2 A and the transformer is ideal, the primary current is 0.2 x 10 = 2 A, and the power is the same on both sides, 440 W.

If the efficiency is 90 per cent and the output power is 440 W, the input power is 440/0.9 = 489 W. The difference, 49 W, appears as heat in the windings and core. For a transformer of 90 per cent efficiency with an input of 1000 W, an output of 900 W is expected, and the sources of the loss are the copper loss, hysteresis, eddy currents and flux leakage.

4.5 One more thing about the power factor

Exercise questions often combine the power factor with a transformer. An appliance that takes 1000 W at a power factor of 0.5 draws an rms current of 1000/(220 x 0.5) = 9.1 A, while the same power at a power factor of one would need only 4.5 A. The extra current heats the supply cables without doing any useful work, which is why large installations add capacitors to correct a lagging power factor. The transformer relation and the power factor together explain why transmission is done at high voltage: the line current falls for the same power, and both the cable loss and the voltage drop shrink.