Inductor and capacitor in an AC circuit

Two parts that treat AC and DC differently

Join a lamp to a battery through a capacitor. The lamp stays dark. Replace the battery with an AC source of the same voltage. Now the lamp glows. Join the lamp to a battery through a coil of many turns on an iron core. The lamp is bright. Replace the battery with the AC source. The lamp is dimmer.

A resistor cannot do this. It gives the same $V/I$ for direct and alternating current. A coil and a capacitor depend on how fast the voltage changes. Let us see how.

A coil in an AC circuit

Take a pure inductor, a coil whose own wire has no resistance. Its inductance is $L$. Join it to the source $v = V_m \sin \omega t$.

When the current in the coil changes, the coil makes an emf $-L\,\dfrac{di}{dt}$ (Faraday's law). The source voltage must balance it at every instant, so

Put in $v = V_m \sin \omega t$. Then $\dfrac{di}{dt} = \dfrac{V_m}{L}\sin \omega t$. The current is the integral of this:

So the peak current is $I_m = \dfrac{V_m}{\omega L}$. The factor $\omega L$ plays the role of a resistance. It is the inductive reactance:

Its unit is the ohm. Check: henry times rad/s is $\dfrac{\text{V s}}{\text{A}} \times \dfrac{1}{\text{s}} = \dfrac{\text{V}}{\text{A}}$. With it, $I_m = V_m/X_L$ and $I_{rms} = V_{rms}/X_L$.

The current lags the voltage by $90^\circ$. The current is greatest when it stops changing, so when $di/dt = 0$. At that instant $v = L\,di/dt = 0$. So the current peak comes a quarter of a cycle after the voltage peak.

No power on average. The power is $p = vi = -\tfrac{1}{2} V_m I_m \sin 2\omega t$. It is positive for a quarter of a cycle and negative for the next quarter. The coil takes energy from the source and stores it in its magnetic field. Then it gives all of it back. The average power is zero.

What if the frequency changes? $X_L$ is in proportion to $f$. At $f = 0$ (direct current) a pure coil has no reactance at all. At a high frequency it is a large opposition. This is why a coil is used as a choke, to hold back fast changes.

A worked example. A pure inductor of $0.25$ H is joined to the $230$ V, $50$ Hz mains. Find the reactance, the rms current and the peak current.

1. The reactance. $X_L = 2\pi f L = 2\pi \times 50\ \text{Hz} \times 0.25\ \text{H} = 78.5\ \Omega$.
2. The rms current. $I_{rms} = \dfrac{V_{rms}}{X_L} = \dfrac{230\ \text{V}}{78.5\ \Omega} = 2.93$ A.
3. The peak current. $I_m = \sqrt{2}\, I_{rms} = 1.414 \times 2.93\ \text{A} = 4.14$ A.

A capacitor in an AC circuit

Now join a capacitor $C$ to the same source. The charge on its plates is $q = Cv = C V_m \sin \omega t$. The current is the rate at which the charge changes:

The peak current is $I_m = \omega C V_m$. Write it as $I_m = V_m/X_C$. The capacitive reactance is

and its unit is also the ohm.

The current leads the voltage by $90^\circ$. The current is greatest when the charge is changing fastest. That is when $v$ passes through zero, a quarter of a cycle before its peak.

No charge crosses the gap. The gap between the plates never conducts. Charge only moves to and fro in the wires, filling and emptying the plates. The lamp in the first example glows because of this to-and-fro current.

No power on average. The power is $p = vi = \tfrac{1}{2} V_m I_m \sin 2\omega t$. The capacitor stores energy in its electric field and gives it back. The average power is zero.

What if the frequency changes? $X_C$ is in inverse proportion to $f$. At $f = 0$ (direct current) it is infinite, so the capacitor blocks direct current. At a high frequency it is small.

A worked example. A capacitor of $20\ \mu$F is joined to the $230$ V, $50$ Hz mains. Find the reactance and the rms current.

1. The reactance. $X_C = \dfrac{1}{2\pi f C} = \dfrac{1}{2\pi \times 50\ \text{Hz} \times 20 \times 10^{-6}\ \text{F}} = 159\ \Omega$.
2. The rms current. $I_{rms} = \dfrac{V_{rms}}{X_C} = \dfrac{230\ \text{V}}{159\ \Omega} = 1.44$ A.
3. The peak current. $I_m = 1.414 \times 1.44\ \text{A} = 2.04$ A.

Raise the frequency. What happens to the current in the coil, and to the current in the capacitor? Which of the two curves is ahead of the voltage curve?

Which one is ahead?

A short phrase helps: ELI the ICE man. In a coil (L) the voltage E comes before the current I, so E, L, I. In a capacitor (C) the current I comes before the voltage E, so I, C, E.

• Resistor: $i$ is in phase with $v$. It turns the energy of the source into heat.
• Coil: $i$ lags $v$ by $90^\circ$. $X_L = \omega L$ grows with the frequency.
• Capacitor: $i$ leads $v$ by $90^\circ$. $X_C = 1/(\omega C)$ falls with the frequency.

Practise with new numbers

Try it in Python

The program joins a coil and a capacitor to a $12$ V peak source, one at a time. It finds each current from the rules $v = L\,di/dt$ and $i = C\,dv/dt$, and then reads off the delay between the current and the voltage.

Common mistakes

• Mixing up lead and lag. The current lags in a coil and leads in a capacitor. Use ELI the ICE man.
• Forgetting the factor $2\pi$. The reactance uses $\omega = 2\pi f$, not $f$.
• Leaving mH or $\mu$F as they are. Change to H and F before using the formulas.
• Thinking charge flows through a capacitor. It flows in the wires and fills and empties the plates.
• Thinking a coil or a capacitor heats the circuit. Their average power is zero. Only resistance turns energy into heat.
• Saying that $X_C$ grows with the frequency. It falls. $X_L$ is the one that grows.

Remember these

• Coil: $X_L = \omega L = 2\pi f L$. The current lags the voltage by $90^\circ$.
• Capacitor: $X_C = \dfrac{1}{\omega C} = \dfrac{1}{2\pi f C}$. The current leads the voltage by $90^\circ$.
• For each: $I_{rms} = \dfrac{V_{rms}}{X}$ and $I_m = \dfrac{V_m}{X}$. The unit of reactance is the ohm.
• A coil passes direct current and holds back fast changes. A capacitor blocks direct current and passes fast changes.
• The average power in a pure coil or a pure capacitor is zero.

Check yourself

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

1. The frequency of the source across a pure coil is doubled. The voltage is the same. What happens to the rms current?

2. A capacitor is joined to a source of $5.0$ V rms. The rms current is $0.020$ A. What is the reactance of the capacitor?

3. The voltage across a capacitor is at its peak. What is the current in the circuit at that instant?

4. A coil of $5.0$ mH is joined first to a $50$ Hz source and then to a $5.0$ kHz source of the same voltage. By what factor does the current change?