Phasors, and the Resistor

An arrow that turns

In the chapter Kinematics of SHM, a point moving round a circle cast a shadow that moved in simple harmonic motion. An alternating voltage is built the same way. Draw an arrow of length \(v_0\) from a fixed point. Let it turn anticlockwise at the angular speed \(\omega\), starting along the horizontal. At time \(t\) it makes the angle \(\omega t\) with the horizontal. The height of its tip is \(v_0\sin\omega t\), which is exactly the voltage at that instant.

In Fig 2.1 the phasor makes one turn in one period, \(20\,\text{ms}\). At \(2.5\,\text{ms}\) it has turned through \(\tfrac{2.5}{20} \times 360° = 45°\). The height of its tip is then \(10\sin 45° = 7.1\,\text{V}\), the value found in the last lesson. At \(5\,\text{ms}\) the phasor points straight up and the voltage is at its peak. At \(10\,\text{ms}\) it points to the left and the voltage is zero again.

Two quantities of the same frequency have phasors that turn together. The angle between them never changes. That fixed angle is their phase difference. A phasor ahead of another, anticlockwise, belongs to the quantity that reaches its peak earlier. This lesson and the next draw every phasor diagram at one frozen instant, since the whole picture turns as one rigid body.

Circuit problems give rms values, not peaks. Every phasor length can be divided by \(\sqrt2\) without changing any angle. So a phasor diagram is often drawn with rms lengths, as in lesson 4. The angles, and so the phases, are the same either way.

The resistor

Connect a resistor \(R\) across a source \(v = v_0\sin\omega t\). A resistor stores no energy and has nothing to delay the current. So Ohm's law, from the chapter Electric Current and Resistance, holds at every instant: \[ \begin{gathered} i = \dfrac{v}{R} = \dfrac{v_0}{R}\sin\omega t = i_0\sin\omega t \\ i_0 = \dfrac{v_0}{R}. \end{gathered} \] The current rises, falls and reverses together with the voltage. The two are in phase. Their phasors point the same way and turn together.

Dividing both peaks by \(\sqrt2\) gives Ohm's law for rms values, \(V_{rms} = I_{rms}R\). So every direct-current formula for a resistor holds with rms values in place of steady ones.

Adding with phasors

Suppose two voltages of the same frequency act in series, \(v_1 = 40\sin\omega t\) and \(v_2 = 30\sin(\omega t + 90°)\). Their sum at each instant is the sum of two projections. The projection of a vector sum is the sum of the projections. So the sum is the projection of the vector sum of the two phasors. That vector sum turns with them, so the total is again a sinusoid of the same frequency.

In Fig 2.2 the two phasors are at right angles. So the resultant is \(\sqrt{30^2 + 40^2} = 50\,\text{V}\), and its angle \(\phi\) from the \(40\,\text{V}\) phasor has \(\tan\phi = 30/40 = 0.75\). That gives \(\phi = 37°\). So \(v_1 + v_2 = 50\sin(\omega t + 37°)\). The peaks do not add to \(70\,\text{V}\), because the two voltages reach their peaks a quarter cycle apart.

Power in a resistor

The power delivered to the resistor at an instant is \(p = vi\). With \(v = v_0\sin\omega t\) and \(i = i_0\sin\omega t\), \[ p = v_0 i_0\sin^2\omega t. \] This is never negative. It is zero twice each cycle, when the voltage and current pass through zero together. It peaks at \(v_0 i_0\) twice each cycle. So on the \(50\,\text{Hz}\) mains the power in a filament pulses \(100\) times a second. The resistor only ever takes energy from the source; it never gives any back.

The mean of \(\sin^2\omega t\) is \(\tfrac12\), so the average power is \(v_0 i_0/2\). Split the \(\tfrac12\) as \(\tfrac{1}{\sqrt2} \times \tfrac{1}{\sqrt2}\), and the average power is \(V_{rms}I_{rms}\). This is why rms values were defined: the direct-current power formulas hold unchanged. The \(100\,\text{W}\) bulb of lesson 1 draws \(0.64\,\text{A}\) at its peak, when the voltage is \(311\,\text{V}\). So its power swings between \(0\) and \(311 \times 0.64 = 200\,\text{W}\), and averages \(100\,\text{W}\).