Worked examples: polar ⇌ rectangular, adding two voltages, and the jω shortcut

Three problems. First: a straight conversion between polar and rectangular form. Second: adding two phasors at different angles to find a resultant (the classic Kirchhoff sum). Third: using the $d/dt \leftrightarrow j\omega$ rule to find the voltage across an inductor without touching calculus.

Example 1 — Convert V = 8 − j6 to polar form

1. Magnitude.
   $$|V| = \sqrt{a^2 + b^2} = \sqrt{8^2 + (-6)^2} = \sqrt{100} = 10$$
2. Angle.
   $$\theta = \arctan(b/a) = \arctan(-6/8) \approx -36.87°$$
   Negative angle because the phasor sits below the real axis (fourth quadrant).
3. Polar form.
   $$V = 10 \angle -36.87°$$
   Sanity check: going back, $10 \cos(-36.87°) = 8$ ✓ and $10 \sin(-36.87°) = -6$ ✓.

Example 2 — V₁ = 10∠0° plus V₂ = 6∠90°

Two voltages at the same frequency. Add them phasorially.

1. Convert both to rectangular.
   $$V_1 = 10 \cos(0°) + j\,10 \sin(0°) = 10 + j0$$
   $$V_2 = 6 \cos(90°) + j\,6 \sin(90°) = 0 + j6$$
   V₂ is a pure imaginary because it sits on the positive imaginary axis.
2. Add component-wise.
   $$V = V_1 + V_2 = (10 + 0) + j(0 + 6) = 10 + j6$$
3. Convert back to polar.
   $$|V| = \sqrt{10^2 + 6^2} = \sqrt{136} \approx 11.66$$
   $$\theta = \arctan(6/10) \approx 30.96°$$
   So $V \approx 11.66 \angle 30.96°\,V$.

Because V₁ and V₂ are 90° apart, the triangle is a right triangle and the magnitudes combine as $|V| = \sqrt{|V_1|^2 + |V_2|^2}$. Any time two phasors are in quadrature, Pythagoras is all you need.

Example 3 — Voltage across a 5 mH inductor carrying 2 sin(2π·1000 t) A

An inductor obeys $v = L\,dI/dt$. Pre-phasors, we'd differentiate. With phasors it's a pure multiplication by $j\omega L$.

1. Identify the current phasor. $i(t) = 2 \sin(2\pi \cdot 1000\,t)$ so $I = 2 \angle 0° \, A$ and $\omega = 2\pi \cdot 1000 \approx 6283$ rad/s.
2. Inductor impedance.
   $$Z_L = j\omega L = j \cdot 6283 \cdot 0.005 \approx j31.4\,\Omega$$
   Magnitude 31.4 Ω, angle +90°.
3. Voltage phasor by ‘Ohm’s law’.
   $$V = Z_L \cdot I = (j31.4)(2 \angle 0°) = j62.8 = 62.8 \angle 90°\,V$$
4. Back to the time domain.
   $$v(t) = 62.8 \sin(2\pi \cdot 1000\,t + 90°)$$
   Voltage peaks a quarter-cycle before the current does — ELI, exactly as predicted in Topic 6.