Phasors and reactance

A shadow that swings

Picture a bicycle wheel that turns at a steady rate, with a small lamp on its rim. A wall stands beside the wheel, and the lamp makes a bright spot on it. As the wheel turns, the spot goes up and down the wall. Its height is the height of the lamp above the axle, which is $r \sin \theta$, with $\theta$ the angle that the spoke has turned through.

That is a sine wave. The up-and-down motion of the spot is the same as an alternating voltage. This is the idea of a phasor.

The phasor

A phasor is an arrow that turns anticlockwise at the angular frequency $\omega$ of the source.

• The length of the arrow is the peak value, $V_m$ or $I_m$.
• The vertical part of the arrow, at the time $t$, is the value of the voltage or the current at that time. For a voltage phasor at the angle $\omega t$, it is $V_m \sin \omega t$.

A current $i = I_m \sin(\omega t + \varphi)$ has its own phasor. It turns at the same $\omega$ as the voltage phasor, so the angle between the two arrows stays fixed. That angle is the phase difference $\varphi$. A phasor diagram is a picture of the arrows at one instant. We draw the voltage along the right, as a reference.

• Resistor. $I$ points the same way as $V$. The phase difference is $0^\circ$.
• Coil. $I$ points $90^\circ$ behind $V$, clockwise from it. The current lags.
• Capacitor. $I$ points $90^\circ$ ahead of $V$, anticlockwise from it. The current leads.

Choose $-90$ and watch the arrows. When the blue arrow points straight up, where is the red arrow? Now choose $+90$.

A phase is a time

The arrows go once round in one period $T$. A phase of $\varphi$ is the fraction $\varphi/360^\circ$ of a full turn, so it is a time delay of

At $50$ Hz the period is $20$ ms. A phase of $90^\circ$ is a quarter of a cycle, which is $5$ ms.

A worked example. In a circuit at $50$ Hz the voltage peak comes $3.0$ ms after the current peak. By how many degrees does the current lead the voltage?

1. The fraction of a cycle. The period is $20$ ms. So the delay is $\dfrac{3.0\ \text{ms}}{20\ \text{ms}} = 0.15$ of a cycle.
2. The angle. $0.15 \times 360^\circ = 54^\circ$.
3. Who leads? The current peak comes first, so the current leads the voltage by $54^\circ$.

A reading like this on an oscilloscope tells you the phase difference. Here each part is alone in its circuit, so the angle is $0^\circ$, $90^\circ$ or $-90^\circ$. When a circuit has several parts, the angle can be anything between.

Adding with arrows

Two sources in series have a total voltage $v_1 + v_2$ at every instant. If the two are in phase, the peak values add. If they are not in phase, their peaks come at different times, and the peak of the sum is smaller than the sum of the peaks.

The arrows do the work for us. Add the phasors as arrows, tip to tail. The length of the total arrow is the peak value of the total voltage. The angle of the total arrow is its phase.

A worked example. A source of $36$ V peak is joined in series with a source of $48$ V peak that is $90^\circ$ ahead of the first. Find the peak value of the total and its phase.

1. Draw the arrows. The $36$ V arrow points right. The $48$ V arrow points up. Put the second at the tip of the first. They make two sides of a right-angled triangle.
2. The length. $V_m = \sqrt{36^2 + 48^2}\ \text{V} = 60$ V. (This is the 3, 4, 5 triangle, scaled by $12$.)
3. The angle. $\tan \varphi = \dfrac{48}{36}$, so $\varphi = 53.1^\circ$ ahead of the first source.

The sum is not $36 + 48 = 84$ V. The two peaks come $90^\circ$ apart, so they never come together.

The program does the same sum. It writes each phasor as a complex number, a length and an angle, and adds them.

Reactance and frequency

We have met three kinds of opposition to an AC current:

• Resistance $R$. It does not depend on the frequency. It turns energy into heat.
• Inductive reactance $X_L = \omega L$. It grows with the frequency.
• Capacitive reactance $X_C = \dfrac{1}{\omega C}$. It falls with the frequency.

All three are in ohm, and each follows an Ohm's law form, $V = IX$. A reactance does not heat anything. It decides how large the current is, and also its phase.

The graph shows the rule for a coil of $25$ mH and a capacitor of $2.2\ \mu$F. Reading it gives a clear picture. At low frequency the coil lets the current through easily, and the capacitor holds it back. At high frequency it is the other way round.

This is why a capacitor is used to pass high tones and block low ones. A capacitor of $6.8\ \mu$F has $X_C = 23.4\ \Omega$ for a tone of $1$ kHz and only $2.34\ \Omega$ for a tone of $10$ kHz. The high tone goes through ten times as easily. A coil does the opposite. It passes the low tones and holds back the high ones.

The program finds the reactance of the coil and of the capacitor of the graph at five frequencies.

Practise with new numbers

Common mistakes

• Adding peak values of voltages that are not in phase. Add the arrows, not the numbers.
• Drawing the current of a coil ahead of the voltage. A coil lags. A capacitor leads.
• Forgetting that the phasors turn. The arrows turn together, so only the angle between them matters.
• Turning an angle into a time without the period. A phase of $\varphi$ is $\dfrac{\varphi}{360^\circ}\, T$, not $\varphi$ divided by $f$.
• Mixing resistance and reactance. Only a resistance turns energy into heat. A reactance changes the size and the phase of the current.
• Mixing the rotation sense. A phasor ahead of another one is drawn anticlockwise from it.

Remember these

• A phasor turns at $\omega$. Its length is the peak value and its vertical part is the value at that instant.
• The angle between two phasors is the phase difference. It stays fixed.
• A phase $\varphi$ is a delay $\Delta t = \dfrac{\varphi}{360^\circ}\, T$. At $50$ Hz, $90^\circ$ is $5$ ms.
• Add AC voltages of different phase as arrows, tip to tail.
• $R$ does not change with the frequency. $X_L = \omega L$ rises with it, and $X_C = \dfrac{1}{\omega C}$ falls with it.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A voltage leads a current by $36^\circ$ at $50$ Hz. By how much time does the voltage peak come before the current peak?

2. Two equal voltages of $20$ V peak are joined in series, with the second $120^\circ$ ahead of the first. What is the peak value of the total?

3. How does the reactance of a capacitor at $5$ kHz compare with its reactance at $50$ Hz?

4. A coil and a capacitor have the same reactance at $800$ Hz. At $1600$ Hz, which has the larger reactance, and by what factor?