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For a certain load true power is 150KW and reactive power is 125KVAr. Find the apparent power.A. 19.5KVAB. 195.2KVAC. 275 KVAD. 25KVA

Last updated date: 20th Jun 2024
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Hint:-Apparent power is the total power available to us .
Throughout the question we will denote apparent power by S, true power by P and reactive power by Q.
Therefore, from the power triangle we have relation :
${S^2} = {P^2} + {Q^2}$
Using this relation we will find the apparent power.

Complete step-by-step solution:

The above figure is called a power triangle which shows us the relation between true power, apparent power and reactive power.
Apparent power: it is total power available to us for use .
True power: it is the consumed power which is used by the consumer.
Reactive power : it is the unused power available .
As per the relationship between true power, reactive power and apparent power from power triangle we have :
${S^2} = {P^2} + {Q^2}$
Let’s do the calculation part now by putting the concerned values in the power equation .
$\Rightarrow {S^2} = {P^2} + {Q^2} \\ \Rightarrow {S^2} = {(150)^2} + {(125)^2} \\ \Rightarrow {S^2} = 22500 + 15625 \\ \Rightarrow {S^2} = 38125 \\ \Rightarrow S = \sqrt {38125} \\ \Rightarrow S = 195.25KVA \\$(done the square of both P ,Q value, added then and took the square root )
Thus option 2 is correct.

Note:-
True power has units kilowatt(KW) , Apparent power has unit kilo volt ampere and reactive power is calculated in kilo volt ampere reactive . The ratio of true power and apparent power must be kept close to 1 for better power consumption without much losses in the system. If the ratio is not kept close to 1 various other equipment like over excited synchronous motors at no load is added to the system ,to improve the ratio close to 1.