Question

A 67 Vrms source is placed across a load that consists of a 12 ohm resistor in series with an capacitor whose reactance is 5 ohms. Compute the following: a) The average power of the load b) The reactive power of the load c) The apparent power of the load d) The power factor of the load

Answer 1

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:

$P=I^2R$

where

I is the current

R is the resistance

In this problem, we have

V = 67 V (rms voltage)

$R=12 \Omega$ is the resistance of the load

$X=5\Omega$ is the reactance of the circuit

First we have to find the impedance of the circuit:

$Z=\sqrt{R^2+X^2}=\sqrt{12^2+5^2}=13 \Omega$

Then we can find the current in the circuit by using Ohm's law:

$I=\frac{V}{Z}=\frac{67}{13}=5.15 A$

Therefore, the average true power is

$P=I^2R=(5.15)^2(12)=318.3 W$

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

$Q=I^2X$

where

I is the current

X is the reactance of the circuit

In this circuit, we have

$I=5.15 A$ (current)

$X=5 \Omega$ (reactance)

Therefore, the reactive power is

$Q=(5.15)^2(5)=132.6W$

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

$S=I^2Z$

where

I is the current

Z is the impedance of the circuit

Here we have

I = 5.15 A (current)

$Z=13 \Omega$ (impedance)

Therefore, the apparent power is

$S=I^2 Z=(5.15)^2(13)=344.8 W$

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:

$PF=\frac{P}{S}$

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

$PF=\frac{318.3}{344.8}=0.92$

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