How to Calculate Complex Power
Complex power S in volt-amperes (VA) is the sum of real power P in Watts (W) and reactive power Q in volt-amperes reactive (VAr). It is found by calculating the real power determined by the resistive elements of a circuit and the reactive power determined by inductive and capacitive elements and placing them in a complex equation S=P+jQ where j is the complex operator, the square root of -1.
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Things You'll Need
- Circuit voltage V in volts, V
- Circuit current A in amps, A
- Circuit frequency F in Hertz, Hz
- Resistance R in Ohms
- Inductance L in Henrys, usually expressed in milli-Henrys, mH
- Capacitance C in Farads, usually expressed in micro-Farads, uF
Instructions
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Determine the real power of the circuit by placing the resistance and the current in the formula P=I^2R where P is the real power in Watts (W), I is the current in amps (A) and R is the resistance in Ohms.
In a real industrial circuit, I might be 10 A, R might be 10 ohms and the real power would be 1000 W.
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Calculate the reactance X of an inductor from the inductance and the frequency in the formula X=2piFL where X is the reactive power, pi is the constant 3.14, F is the frequency in Hz and L is the inductance in Henrys (H).
In a real circuit, F is usually 60 Hz and L might be 100 mH so that X would be 37.68 ohms reactive.
The reactive power Q is found using the formula Q=I^2X. For the circuit with 10 A, the reactive power Q would be 3768 VAr.
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3
Calculate the reactance X of a capacitor is from the capacitance and the frequency in the formula X=1/(2piFC) where X is the reactive power, pi is the constant 3.14, F is the frequency in Hz and C is the capacitance in Farads.
In a real circuit, F is usually 60 Hz and C might be 100 uF so that Q would be 26.54 ohms reactive.
The reactive power Q is found using the formula Q=I^2X. For the circuit with 10 A, the reactive power Q would be 2654 VAr.
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4
Determine the complex power by adding the real and reactive powers in the formula S=P+jQ. The reactive powers of inductances and capacitances are of opposite sign since the voltages and currents are out of phase in opposite directions, so the formula becomes S=P+j(Qcapacitive-Qinductive).
For the sample circuit above, which has the resistor, capacitor and inductor in series, the complex power would be S=1000+j(2654-3768) or S=1000+j1114. For more complicated circuits with parallel, or series/parallel branches, the resistive and reactive elements combine just as resistors in a DC circuit would, so that the overall calculations above remain valid.
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References
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