Phasor Study of AC

From the rotating vector to the sine wave

A vector of length \( V_p \) rotates at constant speed. Its shadow on the vertical axis draws the sine; its shadow on the horizontal axis, the cosine. The sine wave is no accident: it is what you see when something rotates.

sine · vertical projection cosine · horizontal projection rotating vector (phasor)

\[ v (t) = V_p \operatorname{sin} (\omega t + \varphi) \] ω is the angular velocity in rad/s, t is time in seconds and φ is the initial phase: where the vector was when the stopwatch started.

\( \omega = 2 \pi f \) · \( T = \frac{1}{f} \) · \( \text{one turn} = 360^\circ = 2 \pi \, \operatorname{rad} \) One full turn of the vector is one cycle of the wave. At 50 Hz it makes 50 turns per second.

Controls

Peak value \( V_p \) 311 V

Frequency f 50 Hz

Initial phase φ 0°

Animation speed 1×

Draw the sine Draw the cosine

At this instant

Angle \( \theta = \omega t + \varphi \)0° v(t)0 V Cosine projection311 V

ω314 rad/s Period T20 ms V RMS220 V

Watch out for this: the cosine is the same sine advanced by 90°. That is why a single figure —the rotating vector— explains both functions, and why phase shifts are measured in degrees.

The values of a sine wave

An AC signal changes all the time, so a single number cannot describe it. Four are used, and you have to know which one each situation calls for.

v(t) RMS average over a half cycle

Try it

Peak value \( V_p \) 311 V

Value	Formula	Equals
Peak	\( V_p \)	311 V
Peak-to-peak	\( 2 V_p \)	622 V
RMS	\( \frac{V_p}{\sqrt{2}} \)	220 V
Average	\( \frac{2 V_p}{\pi} \)	198 V

The most widely used fact: when we say the mains is 220 V we are talking about the RMS value. The real peak is 220·√2 ≈ 311 V, and peak to peak it is 622 V. That is why a capacitor for the mains is never chosen with a 250 V rating.

Why RMS? It is the DC value that would heat a resistor equally. It is the one a multimeter reads and the one used to calculate power. The average value over a full cycle is zero, because the positive and negative half cycles cancel: that is why a half cycle is used.

How each component behaves

In the resistor the current follows the voltage without lagging or leading: they are in phase.

voltage v(t) current i(t) On the left, the same two vectors rotating together.

Controls

Frequency f 50 Hz

Resistance R 100 Ω

What happens

\[ X = R \]

Opposition to current100 Ω Phase shift0° With V = 220 V, I equals2.2 A

The resistor does not discriminate: it opposes 1 Hz exactly as it opposes 1 MHz. That is why its opposition is called resistance and not reactance, and it does not depend on frequency.

To remember it: “ELI the ICE man.” In the inductor (L) the voltage E comes before the current I; in the capacitor (C) the current I comes before the voltage E.

Impedance of a series RLC circuit

In AC the oppositions do not add head-on: R lies on the real axis and the reactances on the imaginary one, because they are 90° apart. The sum is vectorial and forms a triangle.

R \( X_L \) \( X_C \) Z

How Z changes with frequency

The inductor “closes” as frequency rises and the capacitor “opens.” Where they are equal there is resonance: the impedance drops to R and the current is at its maximum.

Circuit

Source voltage (RMS) 220 V

Frequency f 50 Hz

R 100 Ω

L 300 mH

C 30 µF

Results

\( X_L = 2 \pi f L \)94.2 Ω \( X_C = \frac{1}{2 \pi f C} \)106.1 Ω \( X = X_L -X_C \)−11.9 Ω

\( Z = \sqrt{R^2 + X^2} \)100.7 Ω \( \varphi = \mathrm{arctan} \, \frac{X}{R} \)−6.8° \( I = \frac{V}{Z} \)2.18 A

\( V_R = I R \)218 V \( V_L = I X_L \)206 V \( V_C = I X_C \)232 V \( \sqrt{V_R^2 + {\left(V_L -V_C \right)}^2} \)220 V

Resonance \( f_0 \)53.1 Hz The circuit iscapacitive

What is surprising: \( V_L \) and \( V_C \) can each come out larger than the source and there is no mistake. Since they are 180° apart, they subtract from each other, and the vector sum of the three does give the source voltage.

The plot of V against I

Instead of looking at the two waves against time, you put the voltage on one axis and the current on the other. What the point traces gives the component away: a straight line if it is resistive, an ellipse if there is reactance, a circle if it is pure reactance.

voltage current operating point

Phase shift

φ between v and i 0°

The buttons above move this control by themselves. Move it by hand to see how the line gradually opens up until it becomes a circle.

How to read it

Shapeline \( \varphi \)0° \( \cos \, \varphi \)1.00 Enclosed areazero

The slope: here both axes are normalized, which is why the line for the resistive case always comes out at 45°. On a real oscilloscope the slope equals R: the steeper the line, the larger the resistor.

What the area means: if the shape is a straight line, all the energy that goes in is consumed. The wider the ellipse opens, the more energy goes back and forth between the source and the component without being consumed: that is the reactive power.

On the oscilloscope: this shape appears when you put it in XY mode, with the voltage on one channel and, on the other, the drop across a small resistor that acts as a current sensor.

Instantaneous power \( p (t) = v i \)

Multiply the two waves point by point. When the product is positive the source delivers energy; when it is negative, the component gives it back. What is left over as the net balance is the active power.

The power triangle

Controls

Voltage V (RMS) 220 V

Current I (RMS) 10 A

Phase shift φ 30°

The three powers

Apparent \( S = V I \)2200 VA Active \( P = V I \cos \, \varphi \)1905 W Reactive \( Q = V I \, \operatorname{sin} \, \varphi \)1100 VAr

Power factor \( \cos \, \varphi \)0.87

\[ S^2 = P^2 + Q^2 \]

Why the power company cares: the installation has to handle S (the VA), but the customer is billed for P (the W). With a low cos φ extra current flows that does no work but heats the cables. That is why it is corrected with capacitors in parallel.

Practice

Five exercises with random numbers. Answer with two decimals; a difference of up to 2 % is accepted, so you can round the way you would in your notebook.