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:
[tex]P=I^2R[/tex]
where
I is the current
R is the resistance
In this problem, we have
V = 67 V (rms voltage)
[tex]R=12 \Omega[/tex] is the resistance of the load
[tex]X=5\Omega[/tex] is the reactance of the circuit
First we have to find the impedance of the circuit:
[tex]Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega[/tex]
Then we can find the current in the circuit by using Ohm's law:
[tex]I=\frac{V}{Z}=\frac{67}{13}=5.15 A[/tex]
Therefore, the average true power is
[tex]P=I^2R=(5.15)^2(12)=318.3 W[/tex]
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
[tex]Q=I^2X[/tex]
where
I is the current
X is the reactance of the circuit
In this circuit, we have
[tex]I=5.15 A[/tex] (current)
[tex]X=5 \Omega[/tex] (reactance)
Therefore, the reactive power is
[tex]Q=(5.15)^2(5)=132.6W[/tex]
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
[tex]S=I^2Z[/tex]
where
I is the current
Z is the impedance of the circuit
Here we have
I = 5.15 A (current)
[tex]Z=13 \Omega[/tex] (impedance)
Therefore, the apparent power is
[tex]S=I^2 Z=(5.15)^2(13)=344.8 W[/tex]
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:
[tex]PF=\frac{P}{S}[/tex]
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
[tex]PF=\frac{318.3}{344.8}=0.92[/tex]
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