Measurement of Power Questions and Answers

Answer: D

The two wattmeters read equal when \(\phi = 0^{\circ}\).

In the two wattmeter method with a balanced load, the readings are \(W_{1} = V_{L}I_{L}\cos(30^{\circ} - \phi)\) and \(W_{2} = V_{L}I_{L}\cos(30^{\circ} + \phi)\). They are equal only when \(30^{\circ} - \phi = 30^{\circ} + \phi\), that is at unity power factor, \(\phi = 0\).

At \(\phi = 60^{\circ}\) one wattmeter reads zero, at \(\phi = 90^{\circ}\) the readings are equal and opposite, and at \(\phi = 30^{\circ}\) one reading is twice the other.

Answer: C

The readings are 17.9 kW and 7.1 kW.

In the two-wattmeter method the sum of the readings is the total power, and their difference is related to it through the power factor angle:

\[ \begin{aligned} W_{1} + W_{2} &= 25 \\ W_{1} - W_{2} &= \frac{(W_{1} + W_{2})\tan\varphi}{\sqrt{3}} \\ &= \frac{25 \times 0.75}{1.732} \approx 10.83 \end{aligned} \]

Here \(\tan\varphi = 0.6/0.8 = 0.75\). Solving, \(W_{1} \approx 17.9\) kW and \(W_{2} \approx 7.1\) kW.

Equal readings of 12.5 kW occur only at unity power factor. The pairs 20.2 and 4.8, and 10.9 and 14.1, add up to 25 kW but have the wrong difference for a power factor of 0.8.

Answer: A

The answer is Statement I is false and Statement II is true.

In a low power factor wattmeter the pressure (voltage) coil is usually connected on the load side, so the current coil carries the load current plus the small current of the pressure coil circuit. At low power factor the deflecting torque is small and this extra current causes a noticeable error. A compensating coil, carrying the pressure coil current and wound alongside the current coil so as to oppose it, cancels the field due to that current.

The coil therefore compensates for the effect of the voltage (pressure) coil circuit, as Statement II says, and not for the impedance of the current coil, so Statement I is false.

Answer: A

Two coils are used, a fixed current coil and a moving pressure coil, and the torque is proportional to the product of their currents. Being a product it gives true power on any waveform and at any power factor, which no single-coil movement can do.

Answer: B

The fixed coil is the current coil, wound with few turns of thick wire and connected in series with the load. The moving coil is the pressure coil, of many turns of fine wire with a series resistance, and is connected across the load.

Answer: C

Averaged over a cycle the torque is proportional to \( V I \cos\phi \), which is exactly the active power. The cosine term is what makes the instrument read power rather than volt-amperes, and why it reads correctly at any power factor.

Answer: D

The rate of change of mutual inductance with angle is made nearly constant by the shaping of the coils, so the deflection is proportional to the power and the divisions are equal. This is in contrast with the moving iron and hot wire movements, whose scales are square law.

Answer: A

The current in the pressure coil must be proportional to the voltage and in phase with it, which needs a circuit whose resistance far outweighs its reactance. Keeping the current small also limits the error it introduces into the current coil reading.

Answer: B

Eddy current damping is ruled out, because a permanent magnet near the coils would distort the very field on which the reading depends. A light vane moving in a closed sector supplies the damping without introducing any magnetic disturbance.

Answer: C

The current coil then carries the load current only, and the pressure coil measures the voltage across the load plus the drop in the current coil, so the current coil loss is included. This connection suits a heavy load current at low voltage.