In the RC parallel circuit shown in Figure 23–2, assume that a resistance of 30 Ω is connected in parallel with a capacitive reactance of 20 Ω. The circuit is connected to a voltage of 240 VAC and a frequency of 60 Hz. Calculate the following circuit values: IR—current flow through the resistor

Question

In the RC parallel circuit shown in Figure 23–2, assume that a resistance of 30 Ω is connected in parallel with a capacitive reactance of 20 Ω. The circuit is connected to a voltage of 240 VAC and a frequency of 60 Hz. Calculate the following circuit values:

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

P—watts (true power)

[latex]I_{C}[/latex]—current flow through the capacitor

[latex]VARs_{C}[/latex]—volt-amperes reactive (reactive power)

C—capacitance of the capacitor

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

Z—total impedance of the circuit

VA—volt-amperes (apparent power)

PF—power factor

∠θ—angle theta (phase angle of voltage and current)

Accepted Answer

Resistive Current
The amount of current flow through the resistor ([latex]I_{R}[/latex]) can be calculated by using the formula

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

[latex]I_{R} = \frac{240 V}{30 \Omega}[/latex]

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

True Power
The amount of total true power (P) in the circuit can be determined by using any of the values associated with the pure resistive part of the circuit. In this example, true power is found using the formula

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

[latex]P = 240 V imes 8 A[/latex]

[latex]P = 1920 W[/latex]

Capacitive Current
The amount of current flow through the capacitor ([latex]\pmb{I_{C}}[/latex]) is calculated using the formula

[latex]I_{C} = \frac{E}{X_{C}}[/latex]

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

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

Reactive Power
The amount of reactive power ([latex]VARs_{C}[/latex]) can be found using any of the total capacitive values. In this example, [latex]VARs_{C}[/latex] is calculated using the formula

[latex]VARs_{C} = E imes I_{C}[/latex]

[latex]VARs_{C} = 240 V imes 12 A[/latex]

[latex]VARs_{C} = 2880[/latex]

Capacitance
The capacitance of the capacitor can be calculated using the formula

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

[latex]C = \frac{1}{377 imes 20 \Omega}[/latex]

[latex]C = \frac{1}{7540}[/latex]

[latex]C = 0.0001326 F = 132.6 \mu F[/latex]

Total Current
The voltage is the same across all legs of a parallel circuit. The current flow through the resistor is in phase with the voltage, and the current flow through the capacitor is leading the voltage by 90º (Figure 23–3). The 90º difference in capacitive and resistive current forms a right triangle as shown in Figure 23–4. Because these two currents are connected in parallel, vector addition can be used to find the total current flow in the circuit (Figure 23–5). The total circuit current ([latex]\pmb{I_{T}}[/latex]) flow can be calculated by using the formula:

[latex]I_{T} = \sqrt{I_{R} ^{2} + I_{C} ^{2}}[/latex]

[latex]I_{T} = \sqrt{(8 A)^{2} + (12 A)^{2}}[/latex]

[latex]I_{T} = \sqrt{64 A^{2} + 144 A^{2}}[/latex]

[latex]I_{T} = \sqrt{208 A^{2}}[/latex]

[latex]I_{T} = 14.422 A[/latex]

Impedance
The total circuit impedance (Z) can be found by using any of the total values and substituting Z for R in an Ohm’s law formula. The total impedance of this circuit is calculated using the formula

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

[latex]Z = \frac{240 V}{14.422 A}[/latex]

[latex]Z = 16.641 \Omega[/latex]

The impedance can also be found by adding the reciprocals of the resistance and capacitive reactance. Because the resistance and capacitive reactance are 90º out of phase with each other, vector addition must be used:

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

Another formula that can be used to determine the impedance in a circuit that contains both resistance and capacitive reactance is

[latex]Z = \frac{R imes X_{C}}{\sqrt{R^{2} + X_{C} ^{2}}}[/latex]

Apparent Power
The apparent power (VA) can be calculated by multiplying the circuit voltage by the total current flow:

[latex]VA = E imes I_{T}[/latex]

[latex]VA = 240 V imes 14.422 A[/latex]

[latex]VA = 3461.28[/latex]

Power Factor
The power factor (PF) is the ratio of true power to apparent power. The circuit power factor can be calculated using the formula

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

[latex]PF = \frac{1920}{3461.28}[/latex]

[latex]PF = 0.5547, or 55.47 \%[/latex]

Angle Theta
The cosine of angle theta (∠θ) is equal to the power factor:

cos ∠θ = 0.5547

∠θ = 56.31º

A vector diagram of apparent, true, and reactive power is shown in Figure 23–6. Angle theta is the angle developed between the apparent and true power. The complete circuit with all values is shown in Figure 23–7.

Question 23.1: In the RC parallel circuit shown in Figure 23–2, assume that...

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