Phasors
A rotating arrow for a sine wave
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
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
- 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?
- 311 V
- 269 V
- 220 V
- 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?
- 0 ms: they peak together
- 5 ms
- 10 ms
- 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.