Answer
427.8k+ views
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.
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:
![seo images](https://www.vedantu.com/question-sets/8bed04bc-5d01-4402-9f4e-3e2522952af13043647551961212.png)
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.
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