Waste of energy
Many household electric appliances, for example transformers or motors, have inductive components. The electric utility company does not like to have an inductive load on its distribution network and it requires that a capacitor should be applied so that the inductive component is "compensated". Read more about this issue by searching for "power factor" or "cos phi factor".
To illustrate the advantage of having a compensating capacitor, consider the two electric circuits shown in the Figure. The resistor represents the resistance of the distribution network that is driven by an ideal AC voltage source of voltage . The inductor and the resistor represent the appliance, with a capacitor added for compensation.
We will select numbers corresponding to a situation when the appliance does not use much power. The inductive impedance will be and the load resistance is . The capacitance, if present, will be selected so that the impedance of the whole circuit is purely resistive (does not have a phase shift between the voltage and the current).
Calculate , the power lost in the transmission network for the circuit without the capacitor, and , the power lost in the transmission network when the compensating capacitor is present. What is the ratio , expressed in percentage? (Pick the number closest to your answer.)
(Inspired by Steven Chase's problem . )
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1 solution
We are using complex values, with j = − 1 .
For the first circuit the impedance is Z = R + j ω L = R + j X L (a resitor and inductor in series). The current is I = r + R + j X L V and the power loss is
P 1 = ∣ I ∣ 2 r = V 2 ( r + R ) 2 + X L 2 r .
In the second circuit the impedance is
1 / ( R + j X L ) + j X C 1 = R / ( R 2 + X L 2 ) − j X L / ( R 2 + X L 2 ) + j X C 1 ,
where X C is the impedance of the capacitor. This quantity is real if − j X L / ( R 2 + X L 2 ) 2 + j X C = 0 and then the impedance is a real number, ( R 2 + X L 2 ) / R = R + X L 2 / R . The current is I = r + R + X L 2 / R V and the power loss is
P 2 = ∣ I ∣ 2 r = V 2 ( r + R + X L 2 / R ) 2 r .
The ratio is P 2 / P 1 = ( r + R + X L 2 / R ) 2 ( r + R ) 2 + X L 2 = 0 . 0 0 9 9 0 ≈ 1 % . In fact, with the numbers we have, we can take the limit of r < < X L and R < < X L and this quantity is P 2 / P 1 ≈ X L 2 R 2 = 0 . 0 1 .
With the numbers we have the appliance consumes about 5W of power. Without the capacitor the loss in the network is 0.2W. Although this is not much, with many households the losses add up to a substantial energy waste. The saving is very significant.