If a load is purely inductive and the reactive power is $12$VAR, the apparent power is:
A. $12VA$
B. $6V$
C. $0V$
D. $24V$
Answer
649.2k+ views
Hint: An electric circuit which contains only inductors and no other electrical components like resistors and capacitors is called purely inductive circuit. In a purely inductive circuit the electric circuit lags the voltage applied to the circuit.
Complete step by step solution:
Let the electric potential applied to a purely inductive circuit is $V={{V}_{m}}\sin \omega t$
Then electric current in the circuit is $i={{i}_{m}}\sin \left( \omega t+\dfrac{\pi }{2} \right)$
The instantaneous power of the inductor is given as,
$P=Vi$
\[\begin{align}
& P=\left( {{V}_{m}}\sin \omega t \right)\left( {{i}_{m}}\sin \left( \omega t+\dfrac{\pi }{2} \right) \right) \\
& ={{V}_{m}}{{i}_{m}}\left( \sin \omega t \right)\left( \cos \omega t \right) \\
& =\dfrac{{{V}_{m}}{{i}_{m}}}{2}2\left( \sin \omega t \right)\left( \cos \omega t \right) \\
& \dfrac{{{V}_{m}}}{\sqrt{2}}\dfrac{{{i}_{m}}}{\sqrt{2}}\sin 2\omega t
\end{align}\]
As the average of the sine function over a complete cycle is zero. Therefore, the average power consumed in a purely inductive circuit is zero.
The power which oscillates back and forth that means it moves in both the directions in the circuit or reacts upon itself, is called Reactive Power of the circuit. The reactive power is measured in kilo volt-ampere reactive (kVAR) or Mega-volt-ampere reactive (MVAR).
The given load is purely inductive and the reactive power is 12 VAR.
The combination of reactive power and true power is called apparent power, and it is the product of a circuit's voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.
The product of root mean square (RMS) value of voltage and current is known as Apparent Power. This power is measured in kilo-volt-ampere kVA or Mega-volt-ampere (MVA).
Then the apparent power is 12 VA.
Note:
The average power in a purely capacitive or purely inductive circuit is zero.
The average power of an electric circuit which contains resistance is non-zero.
Complete step by step solution:
Let the electric potential applied to a purely inductive circuit is $V={{V}_{m}}\sin \omega t$
Then electric current in the circuit is $i={{i}_{m}}\sin \left( \omega t+\dfrac{\pi }{2} \right)$
The instantaneous power of the inductor is given as,
$P=Vi$
\[\begin{align}
& P=\left( {{V}_{m}}\sin \omega t \right)\left( {{i}_{m}}\sin \left( \omega t+\dfrac{\pi }{2} \right) \right) \\
& ={{V}_{m}}{{i}_{m}}\left( \sin \omega t \right)\left( \cos \omega t \right) \\
& =\dfrac{{{V}_{m}}{{i}_{m}}}{2}2\left( \sin \omega t \right)\left( \cos \omega t \right) \\
& \dfrac{{{V}_{m}}}{\sqrt{2}}\dfrac{{{i}_{m}}}{\sqrt{2}}\sin 2\omega t
\end{align}\]
As the average of the sine function over a complete cycle is zero. Therefore, the average power consumed in a purely inductive circuit is zero.
The power which oscillates back and forth that means it moves in both the directions in the circuit or reacts upon itself, is called Reactive Power of the circuit. The reactive power is measured in kilo volt-ampere reactive (kVAR) or Mega-volt-ampere reactive (MVAR).
The given load is purely inductive and the reactive power is 12 VAR.
The combination of reactive power and true power is called apparent power, and it is the product of a circuit's voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.
The product of root mean square (RMS) value of voltage and current is known as Apparent Power. This power is measured in kilo-volt-ampere kVA or Mega-volt-ampere (MVA).
Then the apparent power is 12 VA.
Note:
The average power in a purely capacitive or purely inductive circuit is zero.
The average power of an electric circuit which contains resistance is non-zero.
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