Low power factor has disadvantage and demerits in electrical system. It causes voltage fluctuation, requirement of high capacity generator, transformer and other electrical parts within the electrical network from generation side to the consumer side. In this article we shall look into why low power factor is harmful to electrical system.
Additional current is needed with low power factor thus Low Power Factor is Harmful
In a low power factor electrical system, additional current is required. This additional current is itself a loss. Also, it has some other harmful consequences throughout the electrical system. That is the main fact of disadvantage or demerits of low power factor in electrical system.
An example calculation to state Low power factor is harmful
We know electrical power (P) consist of two parts, voltage=V and current=I. Mathematically, it is P=V*I. On the other hand, power factor is an angular parameter between the voltage and current. The value of power factor varies from zero (0) to one (1).
You have heard of low power factor in electrical system. Usual cases, low power factor would be around .5 or .6 etc. High power factor is usually above .90. Ideal power factor is one (1). This is possible with a pure resistive electrical system.
We shall use the basic power formula P (Power)= Voltage (V)*Current (I)* Power factor (PF). As an example, we can choose some easy value, as follows: P=100 Watt, V =200 V. Here, at consumer side, the voltage is 200 V and load is 200 Watt. So, we shall calculate the required current sent from the generator or utility side to run the 200 Watt load.
Case 1: There is pure resistive electrical system at consumer end. Thus power factor is one (1). I=P/(V*PF)=200/(200*1)=1 amp. So, utility side have to transfer 1 amp current to run the 200 watt load at consumer side.
Case 2: There is moderate power factor in electrical system at consumer end. Power factor is .8. I=P/(V*PF)=200/(200*.8)=1.25 amp. So, utility side have to transfer 1.25 amp to run the 200 watt load at consumer side.
Case 3: There is low power factor in electrical system at consumer end. Power factor is .5. I=P/(V*PF)=200/(200*.5)=2 amp. So, utility side have to transfer 2 amp current to run the 200 watt load at consumer side.