## Chapter 25

## Q. 25.1

Assume that the RLC parallel circuit shown in Figure 25–2 is connected to a 240-V, 60-Hz line. The resistor has a resistance of 12 Ω, the inductor has an inductive reactance of 8 Ω, and the capacitor has a capacitive reactance of 16 Ω. Complete the following unknown values:

Z—impedance of the circuit

I_{T}—total circuit current

I_{R}—current flow through the resistor

P—true power (watts)

L—inductance of the inductor

I_{L}—current flow through the inductor

VARs_{L}—reactive power of the inductor

C—capacitance

I_{C}—current flow through the capacitor

VARs_{C}—reactive power of the capacitor

VA—volt-amperes (apparent power)

PF—power factor

∠θ—angle theta

## Step-by-Step

## Verified Solution

**Impedance**

The impedance of the circuit is the reciprocal of the sum of the reciprocals of the legs. Because these values are out of phase with each other, vector addition must be used:

Z = \frac{1}{\sqrt{\left(\frac{1}{R}\right)^{2} + \left(\frac{1}{X_{L}}-\frac{1}{X_{C}}\right)^{2}}}

Z = \frac{1}{\sqrt{\left(\frac{1}{12 \Omega}\right)^{2} + \left(\frac{1}{8 \Omega} – \frac{1}{16 \Omega}\right)^{2}}}

Z = \frac{1}{\sqrt{(0.006944 + 0.003906) \frac{1}{\Omega}}}

Z = \frac{1}{\sqrt{(0.01085 ) \frac{1}{\Omega}}}

Z = \frac{1}{(0.10416) \frac{1}{\Omega}}

Z = 9.601 \Omega

To find the total impedance of the previous example using a scientific calculator, press the following keys. Note that the calculator automatically carries each answer to the maximum number of decimal places. This increases the accuracy of the answer.

\fbox{1} \fbox{2} \fbox{1/x} \fbox{x²} \fbox{+} \fbox{(} \fbox{8} \fbox{1/x} \fbox{-} \fbox{1} \fbox{6} \fbox{1/x } \fbox{)} \fbox{x²} \fbox{=} \fbox{√} \fbox{1/x}

Note that this is intended to illustrate how total parallel resistance can be determined using many scientific calculators. Some calculators may require a different key entry or pressing the equal key at the end.

Another formula that can be used to determine the total impedance of a circuit containing resistance, inductive reactance, and capacitive reactance is

Z = \frac{R \times X}{\sqrt{R^{2} + X^{2}}}

where

X = \frac{X_{L} \times X_{C}}{X_{L} + X_{C}}

In this formula, X_{L} is a positive number and X_{C} is a negative number. Therefore, Z will be either positive or negative depending on whether the circuit is more inductive (positive) or capacitive (negative). To find the total impedance of this circuit using this formula, first determine the value of X:

X = \frac{X_{L} \times X_{C}}{X_{L} + X_{C}}

X = \frac{8 \Omega \times (-16 \Omega)}{8 \Omega + (-16 \Omega)}

X = \frac{-128 \Omega^{2}}{-8 \Omega}

X = 16 \Omega

Now that the value of X has been determined, the impedance can be calculated using the formula

Z = \frac{R \times X}{\sqrt{R^{2} + X^{2}}}

Z = \frac{12 \Omega \times 16 \Omega}{\sqrt{(12 \Omega)^{2} + (16 \Omega)^{2}}}

Z = \frac{192 \Omega}{\sqrt{400 \Omega}}

Z = \frac{192 \Omega}{20 \Omega}

Z = 9.6 \Omega

**Resistive Current**

The next unknown value to be found is the current flow through the resistor. This can be calculated by using the formula

I_{R} = \frac{E}{R}

I_{R} = \frac{240 V}{12 \Omega}

I_{R} = 20 A

**True Power**

The true power, or watts (W), can be calculated using the formula

P = E × I_{R}

P = 240 V × 20 A

P = 4800 W

**Inductive Current**

The amount of current flow through the inductor can be calculated using the formula

I_{L} = \frac{E}{R}

I_{L} = \frac{240 V}{12 \Omega}

I_{L} = 30 A

**Inductive VARs**

The amount of reactive power, or VARs, produced by the inductor can be calculated using the formula

VARs_{L} = E \times I_{L}

VARs_{L} = 240 V \times 30 A

VARs_{L} = 7200

**Inductance**

The amount of inductance in the circuit can be calculated using the formula

L = \frac{X_{L}}{2 \pi f}

L = \frac{8 \Omega}{377}

L = 0.0212 H

**Capacitive Current**

The current flow through the capacitor can be calculated using the formula

I_{C} = \frac{E}{X_{C}}

I_{C} = \frac{240 V}{16 \Omega}

I_{C} = 15 A

**Capacitance**

The amount of circuit capacitance can be calculated using the formula

C = \frac{1}{2 \pi f X_{C}}

C = \frac{1}{377 \times 16 \Omega}

C = 0.000165782 F = 165.782 \mu F

**Capacitive VARs**

The capacitive VARs can be calculated using the formula

VARs_{C} = E \times I_{C}

VARs_{C} = 240 \times 15

VARs_{C} = 3600

**Total Circuit Current**

The amount of total current flow in the circuit can be calculated by vector addition of the current flowing through each leg of the circuit (Figure 25–3). The inductive current is 180º out of phase with the capacitive current. These two currents tend to cancel each other, resulting in the elimination of the smaller and reduction of the larger. The total circuit current is the hypotenuse of the resulting right triangle (Figure 25–4). The following formula can be used to find total circuit current:

I_{T} = \sqrt{I_{R} ^{2} + (I_{L} – I_{C})^{2}}

I_{T} = \sqrt{(20 A)^{2} + (30 A – 15 A)^{2}}

I_{T} = \sqrt{(20 A)^{2} + (15 A)^{2}}

I_{T} = \sqrt{400 A + 225 A}

I_{T} = \sqrt{625 A}

I_{T} = 25 A

The total current could also be calculated by using the value of impedance found earlier in the problem:

I_{T} = \frac{E}{Z}

I_{T} = \frac{240 V}{9.6 \Omega}

I_{T} = 25 A

**Apparent Power**

Now that the total circuit current has been calculated the apparent power, or VAs, can be found using the formula

VA = E × I_{T}

VA = 240 V × 25 A

VA = 6000

The apparent power can also be found by vector addition of the true power and reactive power:

VA = \sqrt{P^{2} + (VARs_{L} – VARs_{C})^{2}}

**Power Factor**

The power factor can now be calculated using the formula

PF = \frac{W}{VA} \times 100

PF = \frac{4800 W}{6000 VA} \times 100

PF = 0.80 \times 100

PF = 80 \%

**Angle Theta**

The power factor is the cosine of angle theta. Angle theta is therefore

cos ∠θ = 0.80

∠θ = 36.87º

The circuit with all calculated values is shown in Figure 25–5.