Assume that the circuit shown in Figure 24–2 has an applied voltage of 240 V at 60 Hz and that the resistor has a value of 12 Ω, the inductor has an inductive reactance of 24 Ω, and the capacitor has a capacitive reactance of 8 Ω. Find the following unknown values: Z—impedance of the circuit

Question

Assume that the circuit shown in Figure 24–2 has an applied voltage of 240 V at 60 Hz and that the resistor has a value of 12 Ω, the inductor has an inductive reactance of 24 Ω, and the capacitor has a capacitive reactance of 8 Ω. Find the following unknown values:

Z—impedance of the circuit

[latex]I_{T}[/latex]—circuit current

[latex]E_{R}[/latex]—voltage drop across the resistor

P—true power (watts)

L—inductance of the inductor

[latex]E_{L}[/latex]—voltage drop across the inductor

[latex]VARs_{L}[/latex]—reactive power of the inductor

C—capacitance

[latex]E_{C}[/latex]—voltage drop across the capacitor

[latex]VARs_{C}[/latex]—reactive power of the capacitor

VA—volt-amperes (apparent power)

PF—power factor

∠θ—angle theta

Accepted Answer

Total Impedance
The impedance of the circuit is the sum of resistance, inductive reactance, and capacitive reactance. Because inductive reactance and capacitive reactance are 180º out of phase with each other, vector addition must be used to find their sum. This method results in the smaller of the two reactive values being subtracted from the larger (Figure 24–3). The smaller value is eliminated, and the larger value is reduced by the amount of the smaller value. The total impedance is the hypotenuse formed by the resulting right triangle (Figure 24–4).

The impedance is calculated by using the formula

[latex]Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}[/latex]

[latex]Z = \sqrt{(12 \Omega)^{2} + (24 \Omega - 8 \Omega)^{2}}[/latex]

[latex]Z = \sqrt{(12 \Omega)^{2} + (16 \Omega)^{2}}[/latex]

[latex]Z = \sqrt{144 \Omega + 256 \Omega}[/latex]

[latex]Z = \sqrt{400 \Omega}[/latex]

[latex]Z = 20 \Omega[/latex]

In the preceding formula, the capacitive reactance is subtracted from the inductive reactance and then the difference is squared. If the capacitive reactance is a larger value than the inductive reactance, the difference is a negative number. The sign of the difference has no effect on the answer, however, because the square of a negative or positive number is always positive. For example, assume that an RLC series circuit contains a resistor with a value of 10 Ω, an inductor with an inductive reactance of 30 Ω, and a capacitor with a capacitive reactance of 54 Ω. When these values are substituted in the previous formula, the difference between the inductive and capacitive reactances is a negative number:

[latex]Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}[/latex]

[latex]Z = \sqrt{(10 \Omega)^{2} + (30 \Omega - 54 \Omega)^{2}}[/latex]

[latex]Z = \sqrt{(10 \Omega)^{2} + (-24 \Omega)^{2}}[/latex]

[latex]Z = \sqrt{(100 + 576 \Omega)^{2}}[/latex]

[latex]Z = \sqrt{676 \Omega}[/latex]

[latex]Z = 26 \Omega[/latex]

Current
The total current flow through the circuit can now be calculated using the formula

[latex]I_{T} = \frac{E_{T}}{Z}[/latex]

[latex]I_{T} = \frac{240 V}{20 \Omega}[/latex]

[latex]I_{T} = 12 A[/latex]

In a series circuit, the current flow is the same at any point in the circuit. Therefore, 12 A flow through each of the circuit components.

Resistive Voltage Drop
The voltage drop across the resistor can be calculated using the formula

[latex]E_{R} = I_{R} imes R[/latex]

[latex]E_{R} = 12 A imes 12 \Omega[/latex]

[latex]E_{R} = 144 V[/latex]

Watts
The true power of the circuit can be calculated using any of the pure resistive values. In this example, true power is found using the formula

[latex]P = E_{R} imes I[/latex]

[latex]P = 144 V imes 12 A[/latex]

[latex]P = 1728 W[/latex]

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

[latex]L = \frac{X_{L}}{2 \pi f}[/latex]

[latex]L = \frac{24 \Omega}{377}[/latex]

[latex]L = 0.0637 H[/latex]

Voltage Drop Across the Inductor
The amount of voltage drop across the inductor can be calculated using the formula

[latex]E_{L} = I imes X_{L}[/latex]

[latex]E_{L} = 12 A imes 24 \Omega[/latex]

[latex]E_{L} = 288 V[/latex]

Notice that the voltage drop across the inductor is greater than the applied voltage.

Inductive VARs
The amount of reactive power of the inductor can be calculated by using inductive values:

[latex]VARs_{L} = E_{L} imes I[/latex]

[latex]VARs_{L} = 288 V imes 12 A[/latex]

[latex]VARs_{L} = 3456[/latex]

Capacitance
The amount of capacitance in the circuit can be calculated by using the formula

[latex]C = \frac{1}{2 \pi f X_{C}}[/latex]

[latex]C = \frac{1}{377 imes 8 \Omega}[/latex]

[latex]C = \frac{1}{3016}[/latex]

[latex]C = 0.000331565 F , or 331.565 \mu F[/latex]

Voltage Drop Across the Capacitor
The voltage dropped across the capacitor can be calculated using the formula

[latex]E_{C} = I imes X_{C}[/latex]

[latex]E_{C} = 12 A imes 8 \Omega[/latex]

[latex]E_{C} = 96 V[/latex]

Capacitive VARs
The amount of capacitive VARs can be calculated using the formula

[latex]VARs_{C} = E_{C} imes I[/latex]

[latex]VARs_{C} = 96 V imes 12 A[/latex]

[latex]VARs_{C} = 1152[/latex]

Apparent Power
The VAs (apparent power) can be calculated by multiplying the applied voltage and the circuit current:

[latex]VA = E_{T} imes I[/latex]

[latex]VA = 240 V imes 12 A[/latex]

[latex]VA = 2880[/latex]

The apparent power can also be found by vector addition of true power, inductive VARs, and capacitive VARs (Figure 24–5). As with the addition of resistance, inductive reactance, and capacitive reactance, inductive VARs, [latex]VARs_{L}[/latex], and capacitive VARs, [latex]VARs_{C}[/latex] are 180º out of phase with each other. The result is the elimination of the smaller and a reduction of the larger. The following formula can be used to determine apparent power:

[latex]VA = \sqrt{P^{2} + (VARs_{L} - VARs_{C})^{2}}[/latex]

[latex]VA = \sqrt{(1728 W)^{2} + (3456 VARs_{L} - 1152 VARs_{C})^{2}}[/latex]

[latex]VA = \sqrt{(1728 W)^{2} + (2304 VARs)^{2}}[/latex]

[latex]VA = \sqrt{8,294,400}[/latex]

[latex]VA = 2880[/latex]

Power Factor
The power factor can be calculated by dividing the true power of the circuit by the apparent power. The answer is multiplied by 100 to change the decimal into a percent:

[latex]PF = \frac{W}{VA} imes 100[/latex]

[latex]PF = \frac{1728 W}{2880 VA} imes 100[/latex]

[latex]PF = 0.06 imes 100[/latex]

[latex]PF = 60 \%[/latex]

Angle Theta
The power factor is the cosine of angle theta:

cos ∠θ = 0.60
∠θ = 53.13º

The circuit with all calculated values is shown in Figure 24–6.

Question 24.1: Assume that the circuit shown in Figure 24–2 has an applied ...

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