Complex Numbers: Real-World Scenarios (6)

hard • 3 min read
Tags Complex Numbers

Question

In AC circuit analysis, voltages and currents are represented as complex numbers (phasors). A circuit has voltage V=10+10iV = 10 + 10i volts and current I=2+iI = 2 + i amperes. Find the impedance Z=V/IZ = V/I and express it in polar form. Also find the average power dissipated.

Solution — Step by Step

Z=VI=10+10i2+iZ = \frac{V}{I} = \frac{10 + 10i}{2 + i}

Multiply numerator and denominator by the conjugate of the denominator:

Z=(10+10i)(2−i)(2+i)(2−i)=(20−10i+20i−10i2)4−i2=20+10i+104+1=30+10i5=6+2iZ = \frac{(10 + 10i)(2 - i)}{(2 + i)(2 - i)} = \frac{(20 - 10i + 20i - 10i^2)}{4 - i^2} = \frac{20 + 10i + 10}{4 + 1} = \frac{30 + 10i}{5} = 6 + 2i

∣Z∣=62+22=40=210|Z| = \sqrt{6^2 + 2^2} = \sqrt{40} = 2\sqrt{10}.

arg⁡(Z)=tan⁡−1(2/6)=tan⁡−1(1/3)≈18.4∘\arg(Z) = \tan^{-1}(2/6) = \tan^{-1}(1/3) \approx 18.4^\circ.

So Z=210∠18.4∘Z = 2\sqrt{10}\angle 18.4^\circ ohms.

In phasor notation, average power is

P=12∣V∣∣I∣cos⁡ϕ=12 Re(V⋅Iˉ)P = \frac{1}{2}|V||I|\cos\phi = \frac{1}{2}\,\text{Re}(V \cdot \bar{I})

where ϕ\phi is the phase angle between VV and II.

∣V∣=200=102|V| = \sqrt{200} = 10\sqrt{2}, ∣I∣=5|I| = \sqrt{5}.

Iˉ=2−i\bar{I} = 2 - i, so VIˉ=(10+10i)(2−i)=30+10iV \bar{I} = (10 + 10i)(2 - i) = 30 + 10i. Real part is 30.

P=30/2=15P = 30/2 = 15 watts.

Why This Works

In AC analysis, complex numbers capture both magnitude and phase of sinusoidal quantities. The impedance Z=V/IZ = V/I is a complex number — its magnitude is the ratio of voltage to current amplitudes, and its argument is the phase difference between voltage and current.

Power dissipation depends on the in-phase component of current (real power), which is exactly Re(VIˉ)/2\text{Re}(V\bar{I})/2. This formula naturally separates real (resistive) power from reactive (inductive/capacitive) power.

Three reflexes for complex-number circuit problems:

  1. Division by complex → multiply by conjugate of denominator.
  2. Polar form → magnitude a2+b2\sqrt{a^2+b^2}, angle tan⁡−1(b/a)\tan^{-1}(b/a).
  3. Average power → Re(VIˉ)/2\text{Re}(V\bar{I})/2.

These three solve 90% of phasor calculations.

Alternative Method

Convert VV and II to polar first:

V=102∠45∘V = 10\sqrt{2}\angle 45^\circ, I=5∠26.57∘I = \sqrt{5}\angle 26.57^\circ.

Then Z=V/I=(102/5)∠(45∘−26.57∘)=210∠18.43∘Z = V/I = (10\sqrt{2}/\sqrt{5}) \angle (45^\circ - 26.57^\circ) = 2\sqrt{10}\angle 18.43^\circ. Same answer.

Students forget that the conjugate is needed for both division and the power formula. Don’t write V/IV/I as VIˉ/∣I∣V\bar{I}/|I| — that’s wrong. The correct form is VIˉ/(IIˉ)=VIˉ/∣I∣2V\bar{I}/(I\bar{I}) = V\bar{I}/|I|^2.

Final answer: Z=6+2i=210∠18.4∘Z = 6 + 2i = 2\sqrt{10}\angle 18.4^\circ ohms; average power =15= 15 W.

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