a) The average true power is 318.3 W
b) The reactive power is 132.6 W
c) The apparent power is 344.8 W
d) The power factor is 0.92
Explanation:
a)
For a circuit made of a resistor and a capacitor, the average (true) power is given by the resistive part of the circuit only.
Therefore, the average true power is given by:
![P=I^2R](https://tex.z-dn.net/?f=P%3DI%5E2R)
where
I is the current
R is the resistance
In this problem, we have
V = 67 V (rms voltage)
is the resistance of the load
is the reactance of the circuit
First we have to find the impedance of the circuit:
![Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega](https://tex.z-dn.net/?f=Z%3D%5Csqrt%7BR%5E2%2BX%5E2%7D%3D%5Csqrt%7B12%5E2%2B5%5E2%7D%3D13%20%5COmega)
Then we can find the current in the circuit by using Ohm's law:
![I=\frac{V}{Z}=\frac{67}{13}=5.15 A](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BV%7D%7BZ%7D%3D%5Cfrac%7B67%7D%7B13%7D%3D5.15%20A)
Therefore, the average true power is
![P=I^2R=(5.15)^2(12)=318.3 W](https://tex.z-dn.net/?f=P%3DI%5E2R%3D%285.15%29%5E2%2812%29%3D318.3%20W)
b)
The reactive power of a circuit consisting of a resistor and a capacitor is the power given by the capacitive part of the circuit.
Therefore, it is given by
![Q=I^2X](https://tex.z-dn.net/?f=Q%3DI%5E2X)
where
I is the current
X is the reactance of the circuit
In this circuit, we have
(current)
(reactance)
Therefore, the reactive power is
![Q=(5.15)^2(5)=132.6W](https://tex.z-dn.net/?f=Q%3D%285.15%29%5E2%285%29%3D132.6W)
c)
In a circuit with a resistor and a capacitor, the apparent power is given by both the resistive and capacitive part of the circuit.
Therefore, it is given by
![S=I^2Z](https://tex.z-dn.net/?f=S%3DI%5E2Z)
where
I is the current
Z is the impedance of the circuit
Here we have
I = 5.15 A (current)
(impedance)
Therefore, the apparent power is
![S=I^2 Z=(5.15)^2(13)=344.8 W](https://tex.z-dn.net/?f=S%3DI%5E2%20Z%3D%285.15%29%5E2%2813%29%3D344.8%20W)
d)
For a circuit with a resistor and a capacitor, the power factor is the ratio between the true power and the apparent power. Mathematically:
![PF=\frac{P}{S}](https://tex.z-dn.net/?f=PF%3D%5Cfrac%7BP%7D%7BS%7D)
where
P is the true power
S is the apparent power
For this circuit, we have
P = 318.3 W (true power)
S = 344.8 W (apparent power)
So, the power factor is
![PF=\frac{318.3}{344.8}=0.92](https://tex.z-dn.net/?f=PF%3D%5Cfrac%7B318.3%7D%7B344.8%7D%3D0.92)
Learn more about power and circuits:
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