Question

In the balanced three-phase circuit shown, \( C_0 = 8.2 \, \mu F \) and the line-to-line r.m.s. voltage is 440 V at 50 Hz. The reading on the wattmeter (in watts) is _________.

Hint:

In a purely capacitive circuit, the wattmeter reads zero because the power factor is zero.

Answer 1

Correct Answer: 141.8

Given: - \( C_0 = 8.2 \, \mu F = 8.2 \times 10^{-6} \, F \) - Line-to-line r.m.s. voltage \( V_{LL} = 440 \, V \) - Frequency \( f = 50 \, Hz \) Step 1: Calculate the impedance of the capacitor
The reactance of a capacitor is given by the formula: \[ X_C = \frac{1}{2\pi f C_0}. \] Substitute the values: \[ X_C = \frac{1}{2 \pi \times 50 \times 8.2 \times 10^{-6}} = 388.23 \, \Omega. \] Step 2: Calculate the current in the circuit
For a balanced three-phase system, the line-to-line voltage is related to the phase voltage \( V_{ph} \) by: \[ V_{ph} = \frac{V_{LL}}{\sqrt{3}} = \frac{440}{\sqrt{3}} = 254.03 \, V. \] The total impedance \( Z \) of the circuit is the combination of the impedance of the capacitor and the resistive element (the wattmeter). For simplicity, assume the total impedance is \( Z = X_C \). The current \( I \) in the circuit is given by Ohm’s law: \[ I = \frac{V_{ph}}{Z} = \frac{254.03}{388.23} = 0.654 \, A. \] Step 3: Calculate the power using the wattmeter reading
The power (in watts) is given by: \[ P = V_{ph} I \cos \phi, \] where \( \cos \phi \) is the power factor. Since this is a purely reactive circuit with a capacitor, \( \cos \phi = 0 \), which means no real power is dissipated. Thus, the reading on the wattmeter is approximately \( 0 \, W \).