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Phasors

NCERT Class 12 Physics, Chapter 7: Alternating Current, §7.3. A short visual lesson: 2 narrated 3D videos, 1 interactive simulation and 2 quiz questions.

A rotating arrow for a sine wave

Narrated video. Transcript:

Drawing sine curves for every voltage and current gets messy. There is a neater way to show both the size and the timing of an ac quantity: the phasor.

A phasor is a vector that rotates about the origin with angular speed ω. Its length is the amplitude, the peak value, of the quantity it stands for.

Draw the voltage phasor V with length v m. At time t, it has turned anticlockwise through an angle ωt from the x-axis.

Now drop a line from its tip onto the vertical axis. The projection is v m sin ωt. That is exactly the instantaneous voltage v.

As the phasor turns, its projection slides up and down. Carry that height across to a graph against time, and it traces out the sine curve of v.

Draw a second phasor I for the current, with length i m. Its projection on the vertical axis gives the instantaneous current i.

For a resistor, v equals v m sin ωt and i equals i m sin ωt. Both phasors point the same way at every instant and turn together. That is what in phase looks like.

Example: v m is 311 volts, and at this instant the phasor makes 30° with the x-axis. The instantaneous voltage is 311 sin 30°, which is 155.5 volts.

Remember, phasors are a picture, not real arrows in space. Only their projections on the vertical axis are the actual voltage and current.

Reading phase from phasors

Narrated video. Transcript:

In many circuits the current is not in phase with the voltage. Then the two phasors are separated by a fixed angle φ, and they keep that angle as they rotate together.

Here the current is i equals i m sin of ωt plus φ. Its phasor is φ ahead of the voltage phasor, in the anticlockwise direction. We say the current leads the voltage by φ.

On the graph, a leading current reaches its peak earlier. If the current phasor were behind, clockwise from V, the current would lag and peak later.

An angle of 360° is one whole period T. So a phase difference of 90°, a quarter turn, means a time gap of T over 4.

At 50 hertz, T is 20 milliseconds. A current lagging by 90° peaks 5 milliseconds after the voltage.

Phasors of the same frequency can be added just like vectors. Take v one equals 3 sin ωt and v two equals 4 sin of ωt plus 90°, both in volts.

Their phasors are at right angles. Add them head to tail: the resultant has length √(3² plus 4²), which is 5 volts.

It is ahead of the first phasor by an angle whose tangent is 4 over 3, about 53°. So v one plus v two equals 5 sin of ωt plus 53°.

Adding two sine curves point by point would take ages. With phasors it took one right triangle. We will use exactly this idea for the series LCR circuit.

Try it: phasors and graphs

Interactive simulation. Things to try:

  • Press Pause, then drag inside the circle to turn the phasors by hand. The dashed lines carry each projection across to the graph.
  • Turn to ωt = 30° and read “v now”: 311 × sin 30° ≈ 156 V.
  • Switch to “L only”: the current phasor now trails the voltage phasor by 90°. On the graph, i peaks a quarter period later.
  • Switch to “C only”: now the current phasor is 90° ahead, so i peaks first.

Projection of a phasor

The voltage phasor has length v_m = 311 V. At one instant it makes an angle of 30° with the x-axis. What is the instantaneous voltage?

  1. 311 V
  2. 269 V
  3. 220 V
  4. 155.5 V
Answer

D. The instantaneous value is the projection on the vertical axis: v = v_m sin ωt = 311 × sin 30° = 311 × 0.5 = 155.5 V. (269 V uses cos 30°, the projection on the wrong axis. 311 V is the peak and 220 V the rms value; neither is the value at this instant.)

Lagging by a quarter turn

In a 50 Hz circuit, the current phasor is 90° behind (clockwise from) the voltage phasor. How long after the voltage reaches its peak does the current reach its peak?

  1. 0 ms: they peak together
  2. 5 ms
  3. 10 ms
  4. 20 ms
Answer

B. One period is T = 1/ν = 1/50 s = 20 ms, and a full turn of 360° takes T. A lag of 90° is a quarter turn, so the current peaks T/4 = 5 ms later. (10 ms would be 180°: the two would be exactly opposite.)

Summary

  • Instantaneous value: v = v_m sin ωt (projection on the vertical axis)
  • Leading current: i = i_m sin(ωt + φ)
  • Phase and time: Δt = (φ/360°) × T
  • A phasor is a vector of length equal to the amplitude, rotating anticlockwise with angular speed ω.
  • Its projection on the vertical axis gives the instantaneous value of the voltage or current.
  • The angle between two phasors is their phase difference. The one further anticlockwise leads.
  • For a resistor, the V and I phasors point the same way: voltage and current are in phase.
  • Phasors of the same frequency add like vectors.