Answer
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Hint: The power factor is $1$ (perfect) for the purely resistive circuit because the reactive power approaches zero. For the completely inductive circuit, the power factor is zero because true power approaches zero. A similar could be stated for a purely capacitive circuit. If there are no resistive components in the system, then the true power must be equivalent to zero, making any strength in the circuit purely reactive.
Complete step-by-step solution:
Power factor is an essential parameter for the estimation of active and reactive power in an electrical circuit. It has importance only for AC circuits. For DC circuits, it is not recognized for power calculation. Its value is one for the DC circuit and may change from zero to one for the AC circuit.
We know that the power used in a DC circuit is given by the product of voltage (V) and current (I). This mean,
$Power = VI$
However, for the AC circuit, the above formula is not right. Another parameter, called the power factor, is also involved. Thus, for the AC circuit,
$Power = VI \times (pf)$, where pf stands for the power factor.
$pf = cos \theta$
In a purely resistive circuit, the current passing through the resistance is always in phase with the implemented voltage. This means that the angle between voltage and current is $0^{\circ}$ and hence
$pf = cos 0 = 1$. This indicates that resistance will only utilize active power. It will not use any reactive power.
In a purely inductive circuit, the current by the inductor lags the used voltage by $90^{\circ}$. Therefore, the angle between voltage and current is $90$ degree, and hence
$pf = cos 90 = 0$ Thus, an inductor will not use any active power. It only uses reactive power.
Similarly, the current through the capacitor leads the voltage by $90^{\circ}$. Therefore
$pf = cos 90 = 0$. In a purely capacitive circuit, a capacitor is thus a generator of reactive power. It does not use any active power.
Note:Power factor can be an essential aspect to consider in an AC circuit because any power factor less than one indicates that the circuit’s cables have to provide more current than what would be required with zero reactance in the circuit to produce the same amount of (true) power to the resistive capacity.
Complete step-by-step solution:
Power factor is an essential parameter for the estimation of active and reactive power in an electrical circuit. It has importance only for AC circuits. For DC circuits, it is not recognized for power calculation. Its value is one for the DC circuit and may change from zero to one for the AC circuit.
We know that the power used in a DC circuit is given by the product of voltage (V) and current (I). This mean,
$Power = VI$
However, for the AC circuit, the above formula is not right. Another parameter, called the power factor, is also involved. Thus, for the AC circuit,
$Power = VI \times (pf)$, where pf stands for the power factor.
$pf = cos \theta$
In a purely resistive circuit, the current passing through the resistance is always in phase with the implemented voltage. This means that the angle between voltage and current is $0^{\circ}$ and hence
$pf = cos 0 = 1$. This indicates that resistance will only utilize active power. It will not use any reactive power.
In a purely inductive circuit, the current by the inductor lags the used voltage by $90^{\circ}$. Therefore, the angle between voltage and current is $90$ degree, and hence
$pf = cos 90 = 0$ Thus, an inductor will not use any active power. It only uses reactive power.
Similarly, the current through the capacitor leads the voltage by $90^{\circ}$. Therefore
$pf = cos 90 = 0$. In a purely capacitive circuit, a capacitor is thus a generator of reactive power. It does not use any active power.
Note:Power factor can be an essential aspect to consider in an AC circuit because any power factor less than one indicates that the circuit’s cables have to provide more current than what would be required with zero reactance in the circuit to produce the same amount of (true) power to the resistive capacity.
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