In this circuit, one resistor is connected in parallel with two inductors (Figure 19–11). The frequency is 60 Hz. The circuit has an apparent power of 6120 VA, the resistor has a resistance of 45 Ω, the first inductor has an inductive reactance of 40 Ω, and the second inductor has an inductive

Question

In this circuit, one resistor is connected in parallel with two inductors (Figure 19–11). The frequency is 60 Hz. The circuit has an apparent power of 6120 VA, the resistor has a resistance of 45 Ω, the first inductor has an inductive reactance of 40 Ω, and the second inductor has an inductive reactance of 60 Ω. It is assumed that both inductors have a Q greater than 10 and their resistance is negligible. Find the following missing values:

Z—total circuit impedance

[latex]I_{T}[/latex]—total circuit current

[latex]E_{T}[/latex]—applied voltage

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

[latex]I_{R}[/latex]—current flow through the resistor

P—watts (true power)

[latex]E_{L1}[/latex]—voltage drop across the first inductor

[latex]I_{L1}[/latex]—current flow through the first inductor

[latex]VARs_{L1}[/latex]—reactive power of the first inductor

[latex]L_{1}[/latex]—inductance of the first inductor

[latex]E_{L2}[/latex]—voltage drop across the second inductor

[latex]I_{L2}[/latex]—current flow through the second inductor

[latex]VARs_{L2}[/latex]—reactive power of the second inductor

[latex]L_{2}[/latex]—inductance of the second inductor

∠θ—the angle that voltage and current are out of phase with each other

Accepted Answer

Impedance
Before it is possible to calculate the impedance of the circuit, the total amount of inductive reactance for the circuit must be found. Because these two inductors are connected in parallel, the reciprocal of their inductive reactances must be added. This will give the reciprocal of the total inductive reactance:

[latex]\frac{1}{X_{LT}} = \frac{1}{X_{L1}} + \frac{1}{X_{L2}}[/latex]

To find the total inductive reactance, take the reciprocal of both sides of the equation:

[latex]X_{LT} = \frac{1}{\frac{1}{X_{L1}} + \frac{1}{X_{L2}}} [/latex]

Refer to the formulas for pure inductive circuits shown in the AC formulas section of the Appendix B. Numeric values can now be substituted in the formula to find the total inductive reactance:

[latex]X_{LT} = \frac{1}{\frac{1}{40 \Omega} + \frac{1}{60 \Omega}} [/latex]

[latex]X_{LT} = \frac{1}{(0.025 + 0.01667) \frac{1}{\Omega}} [/latex]

[latex]X_{LT} = 24 \Omega[/latex]

Now that the total amount of inductive reactance for the circuit is known, the impedance can be calculated using the formula:

[latex]Z = \frac{1}{\sqrt{\left(\frac{1}{R} ight)^{2} + \left(\frac{1}{X_{LT}} ight)^{2}}}[/latex]

[latex]Z = \frac{1}{\sqrt{\left(\frac{1}{45 \Omega} ight)^{2} + \left(\frac{1}{24 \Omega} ight)^{2}}}[/latex]

[latex]Z = 21.176 \Omega [/latex]

A diagram showing the relationship of resistance, inductive reactance, and impedance is shown in Figure 19–12.

[latex]\pmb{E_{T}}[/latex]
Now that the circuit impedance and the apparent power are known, the applied voltage can be calculated using the formula:

[latex]E_{T} = \sqrt{VA imes Z}[/latex]

[latex]E_{T} = \sqrt{6120 VA imes 21.176 \Omega}[/latex]

[latex]E_{T} = 359.996 V[/latex]

[latex]\pmb{E_{R}, E_{L1}, E_{L2}}[/latex]
In a parallel circuit, the voltage must be the same across any leg or branch. Therefore, 360 V is dropped across the resistor, the first inductor, and the second inductor:

[latex]E_{R} = 359.996 V[/latex]

[latex]E_{L1} = 359.996 V[/latex]

[latex]E_{L2} = 359.996 V[/latex]

[latex]\pmb{I_{T}}[/latex]
The total current of the circuit can now be calculated using the formula:

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

[latex]I_{T} = \frac{359.996 V}{21.176 \Omega}[/latex]

[latex]I_{T} = 17 A[/latex]

The remaining values of the circuit can be found using Ohm’s law. Refer to the Resistive-Inductive Parallel Circuits listed in the AC formula section in Appendix B.

[latex]\pmb{I_{R}}[/latex]

[latex]I_{R} = \frac{E_{R}}{R}[/latex]

[latex]I_{R} = \frac{359.996 V}{45 \Omega}[/latex]

[latex]I_{R} = 8 A[/latex]

P

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

[latex]P = 359.996 V imes 8 Ω[/latex]

[latex]P = 2879.968 W [/latex]

[latex]\pmb{I_{L1}}[/latex]

[latex]I_{L1} = \frac{E_{L1}}{X_{L1}}[/latex]

[latex]I_{L1} = \frac{359.996 V}{40 \Omega}[/latex]

[latex]I_{L1} = 9 A[/latex]

[latex]\pmb{VARs_{L1}}[/latex]

[latex]VARs_{L1} = E_{L1} imes I_{L1}[/latex]

[latex]VARs_{L1} = 359.996 V imes 9 A[/latex]

[latex]VARs_{L1} = 3239.964[/latex]

[latex]\pmb{L_{1}}[/latex]

[latex]L_{1} = \frac{X_{L1}}{2 \pi f}[/latex]

[latex]L_{1} = \frac{40 \Omega}{377}[/latex]

[latex]L_{1} = 0.106 H[/latex]

[latex]\pmb{I_{L2}}[/latex]

[latex]I_{L2} = \frac{E_{L2}}{X_{L2}}[/latex]

[latex]I_{L2} = \frac{359.996 V}{60 \Omega}[/latex]

[latex]I_{L2} = 6 A[/latex]

[latex]\pmb{VARs_{L2}}[/latex]

[latex]VARs_{L2} = E_{L2} imes I_{L2}[/latex]

[latex]VARs_{L2} = 359.996 imes 6[/latex]

[latex]VARs_{L2} = 2159.976[/latex]

[latex]\pmb{L_{2}}[/latex]

[latex]L_{2} = \frac{X_{L2}}{2 \pi f}[/latex]

[latex]L_{2} = \frac{60 \Omega}{377}[/latex]

[latex]L_{2} = 0.159 H[/latex]

PF

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

[latex]PF = \frac{2879.968 W}{6120 VA}[/latex]

[latex]PF = 47.06 \%[/latex]

∠θ

cos ∠θ = PF

cos ∠θ = 0.4706

∠θ = 61.93º

A vector diagram showing angle theta is shown in Figure 19–13. The vectors used are those for apparent power, true power, and reactive power. The phase relationship of voltage and current for this circuit is shown in Figure 19–14, and the circuit with all completed values is shown in Figure 19–15.

Question 19.1: In this circuit, one resistor is connected in parallel with ...