Introduction to Electrical Engineering  Basic vocational knowledge (Institut für Berufliche Entwicklung, 213 p.) 
7. Alternating Current 

When loads carry current, a voltage drop is caused. The product of the instantaneous values of current and voltage is called instantaneous power. Normally, the instantaneous power changing from time to time is of less interest than its mean value.
In an effective resistance, current and voltage are in phase. The electrical power becomes completely, i.e. effectively, utilisable. It is called effective power P_{e} (active power). It can be determined on the basis of the effective values according to the relations derived in Section 4.1.
[P_{e}] = W (watt)
In a reactance (ideal coil, ideal capacitor), current and voltage are subjected to a phase shift of 90°. The electrical power is required for the duration of a quarter of a cycle for the building up of the magnetic field (in a coil) or of the electrical field (in a capacitor) and delivered in the subsequent quarter of a cycle. There is no power conversion in a temporal mean. This power is called reactive power P_{r}.
[P_{r}] = Var (voltampere  reactive)
In an impedance, the phase shift between current and voltage is between 0° and 90°. The electrical power is converted partly as effective power and partly as reactive power. This power is called apparent power P_{a}.
[P_{a}] = VA (voltampere)
Since effective resistances and reactances are made up at right angles, this analogously applies to the powers. Consequently, effective and reactive powers are the legs and the apparent power is the hypotenuse of a rightangled triangle having the phase angle (Fig. 7.17.).
Fig. 7.17. Correlation between
effective power, reactive power and apparent power
P_{w} = P_{e}, P_{b} = P_{r}, P_{s} = P_{a}
From this follows
Pa = U · I
where:
Pa 
apparent power 
U 
effective value of voltage 
I 
effective value of current 
From Fig. 7.17. we can cerive
_{}
P_{e} = P_{a} cos j
P_{r} = P_{a} sin j
The expression cos in equation (7.16.a) is called power factor.
From equation (7.16.a) we have
cos j = P_{e}/P_{a}
The power factor cos can be between 0 and 1;
_{}
With a low power factor, the reactive power is high. Since reactive power unnecessarily loads the generators of the power stations and the distribution network, the reactive power must be kept as small as possible for economical reasons; in other words, cos j must be as high as possible. A power factor of cos j = 1 is an ideal case.
In networks of power electrical engineering, an inductive phase shift occurs always because of the necessary transformers and connected motors; this phase shift always worsens the power factor.
The power factor can be improved up to the value of cos j = 1 by means of an additional capacitive component. In practice, capacitors are connected in parallel having a total capacity of
C = P_{r}/(w·U^{2})
where
C 
capacity required to attain a cos j = 1 
P_{r} 
reactive power 
w 
angular frequency 
U 
alternating voltage (effective value) 
Of the energy or work has to be determined., then the product of power times time must be formed according to Section 4.1. In accordance with the various types of power, there are effective work, reactive work and apparent work.
Example 7.6.
An enterprise is connected to a 380 V network (50 Hz). A current of 66 A passes through the loads with a power factor of cos j = 0.5. Determine the apparent power, effective power and reactive power and the capacity of the capacitor necessary to improve the power factor to cos j = 1!
Given:
U = 380 V
I = 66 A
f = 50 Hz
cos j = 0.5
To be found:
P_{a}, P_{e}, P_{r}, C
Solution:
P_{a} = U I
P_{a} = 380 V · 66 A = 25,000 VA
P_{a} = 25 kVA
P_{e} = P_{a} cos j
P_{e} = 25 kVA · 0.5
P_{e} = 12.5 kW
P_{r} = P_{a} sin j
_{}
P_{r} = 25 KVA · 0.865
P_{r} = 21.6 kVar
Proof with equation (7.15.)
_{}
_{}C = P_{r}/(wU^{2})
_{}C » 500 µF
With a capacitor connected in parallel with the loads having a capacity of C 500 µF, the bad power factor of cos j = 0.5 can be increased to the ideal value of cos j = 1.
With alternating current, there are three types of powers, namely, apparent power, effective power and reactive power. Effective power and reactive power are made up at right angles. The apparent power is always greater than effective power or reactive power but it is always smaller than the algebraic sum of the latter two. Efforts are always made to achieve a high effective power and a reactive power which is as small as possible. The ratio of effective power to apparent power is called power factor. It can reach the value of 1 in the most favourable case. The power factor can be improved when a capacitor is connected in parallel with the loads.
Questions and Problems:
1. What are the differences between effective power, reactive power and apparent power?2. What are the values of the effective power and reactive power when the apparent power is 23 kVA with a phase angle of j = 30°?
3. Why is a high power factor desired in energy supply systems? Explain measures by means of which the power can be improved!
4. Two motors are connected, in parallel and to a supply system of 220 V. One motor has a power factor of 0.65 and carries a current of 2 A; the other motor having a power factor of 0.85 carries a current of 5.5 A. Calculate for the total circuit the effective power, reactive power and apparent power and the total power factor!