The series LCR circuit

One loop, three parts

A radio, a tube light and a phone charger all have a resistor, a coil and a capacitor somewhere inside. You have met each part on its own in an AC circuit. Now we put all three in one loop with one source. What current flows, and when does it peak?

The answer starts with one plain fact. In a series circuit there is only one path. At every instant the same current passes the resistor, the coil and the capacitor. So we take the current as our starting point. Then we ask what voltage each part needs.

What each part needs

Let the peak current be $I_0$. Here is what each part does with it.

• Resistor. Its voltage is $V_R = I_0R$. It is in step with the current.
• Coil. Its voltage is $V_L = I_0X_L$, with $X_L = \omega L$. It is a quarter of a cycle, $90^\circ$, ahead of the current.
• Capacitor. Its voltage is $V_C = I_0X_C$, with $X_C = \dfrac{1}{\omega C}$. It is $90^\circ$ behind the current.

The reactances $X_L$ and $X_C$ are measured in ohm. A short way to remember the two angles: in a coil the voltage leads, and in a capacitor the current leads.

Adding three voltages that peak at different times

At every instant the loop rule holds: the source voltage is $v = v_R + v_L + v_C$. But the peak of $v_L$ comes a quarter of a cycle before the peak of $v_R$. So we cannot add the peak values like plain numbers.

We add arrows instead. Each voltage is a phasor, an arrow whose length is the peak value. All the arrows turn together, so we may draw them once, as a frozen picture.

The current is our reference, so $V_R$ points along it. The arrow $V_L$ points $90^\circ$ ahead, which is up in the picture. The arrow $V_C$ points $90^\circ$ behind, which is down. These two point opposite ways, so they partly cancel. What is left is $V_L - V_C$, at a right angle to $V_R$.

The source voltage is the long side of a right triangle:

Now put in $V_R = I_0R$, $V_L = I_0X_L$ and $V_C = I_0X_C$. The current $I_0$ is in every term, so it comes out:

The root is the impedance of the circuit, in ohm:

So $V_0 = I_0Z$, which is Ohm's law again. It holds for rms values too, $V_{rms} = I_{rms}Z$. The angle in the triangle is the phase between the source voltage and the current:

Take care here. The impedance is not $R + X_L + X_C$. The three sides are not in a line. Two of them are at a right angle to each other, and two point opposite ways.

There is a second way to the same result. Write the loop rule as an equation for the charge $q$ on the capacitor, with $i = dq/dt$:

Its steady solution is $i = I_0\sin(\omega t - \phi)$, with the same $Z$ and the same $\phi$. The arrows reach the answer faster, so we use them.

Inductive, capacitive or resistive

We write the current as $i = I_0\sin(\omega t - \phi)$. Then $\phi$ is the angle by which the current lags the voltage. A negative $\phi$ means the current leads.

• $X_L > X_C$. Then $\phi$ is positive and the current lags. The circuit behaves like a coil. We call it inductive.
• $X_L < X_C$. Then $\phi$ is negative and the current leads. The circuit behaves like a capacitor. We call it capacitive.
• $X_L = X_C$. Then $\phi = 0$ and $Z = R$. The circuit behaves like a plain resistor. This is resonance, the subject of the next lesson.

The frequency decides which part wins. The reactance $X_L = \omega L$ grows as the frequency rises. The reactance $X_C = 1/\omega C$ falls. At a low frequency the capacitor wins. At a high frequency the coil wins. So one circuit can be capacitive at one frequency and inductive at another.

A worked example

A source of peak voltage $130$ V and angular frequency $400$ rad/s drives a series circuit. It has $R = 120\ \Omega$, $L = 0.25$ H and $C = 50\ \mu\text{F}$. Find the peak current and the phase.

1. The coil. $X_L = \omega L = 400\ \text{rad/s} \times 0.25\ \text{H} = 100\ \Omega$.
2. The capacitor. $X_C = \dfrac{1}{\omega C} = \dfrac{1}{400\ \text{rad/s} \times 50 \times 10^{-6}\ \text{F}} = 50\ \Omega$.
3. The impedance. $X_L - X_C = 50\ \Omega$. So $Z = \sqrt{(120\ \Omega)^2 + (50\ \Omega)^2} = 130\ \Omega$.
4. The peak current. We want $I_0$, so $I_0 = \dfrac{V_0}{Z} = \dfrac{130\ \text{V}}{130\ \Omega} = 1.0$ A.
5. The phase. $\tan\phi = \dfrac{50\ \Omega}{120\ \Omega} = 0.417$, so $\phi = 22.6^\circ$. It is positive, so the circuit is inductive and the current lags.

Check with the voltages. $V_R = 1.0\ \text{A} \times 120\ \Omega = 120$ V. $V_L = 1.0\ \text{A} \times 100\ \Omega = 100$ V. $V_C = 1.0\ \text{A} \times 50\ \Omega = 50$ V. Then $\sqrt{(120\ \text{V})^2 + (100\ \text{V} - 50\ \text{V})^2} = 130$ V. It matches the source.

Notice that $V_L$ and $V_R$ are each smaller than $V_0$ here, but $V_R + V_L = 220$ V is much more than $V_0$. The peaks do not happen together.

Start with the frequency at its lowest. Which way does the red side point? Raise the frequency step by step. At what frequency does the red side turn round?

Does the current lead or lag?

The mains supply is $230$ V (rms) at $50$ Hz. It feeds a series circuit with $R = 60\ \Omega$, $L = 0.30$ H and $C = 45\ \mu\text{F}$.

1. The angular frequency. $\omega = 2\pi f = 2\pi \times 50\ \text{Hz} = 314\ \text{rad/s}$.
2. The reactances. $X_L = \omega L = 94.2\ \Omega$. $X_C = \dfrac{1}{\omega C} = \dfrac{1}{314\ \text{rad/s} \times 45 \times 10^{-6}\ \text{F}} = 70.7\ \Omega$.
3. The impedance. $X_L - X_C = 23.5\ \Omega$. So $Z = \sqrt{(60\ \Omega)^2 + (23.5\ \Omega)^2} = 64.4\ \Omega$.
4. The rms current. $I_{rms} = \dfrac{230\ \text{V}}{64.4\ \Omega} = 3.57$ A.
5. The phase. $\tan\phi = \dfrac{23.5\ \Omega}{60\ \Omega}$, so $\phi = 21.4^\circ$. The current lags.

What if the frequency is halved to $25$ Hz? Then $X_L$ halves to $47.1\ \Omega$ and $X_C$ doubles to $141.5\ \Omega$. The net reactance is $-94.4\ \Omega$. The circuit is now capacitive, with $Z = 112\ \Omega$, $I_{rms} = 2.06$ A and $\phi = -57.5^\circ$. The current now leads the voltage.

Choose $R = 40\ \Omega$ and a net reactance of $0$. The two waves lie on top of each other. Now make the net reactance $+100\ \Omega$ and then $-100\ \Omega$. Which wave moves ahead in each case?

Practise with new numbers

Try it in Python

The program takes one circuit on the $230$ V mains and finds its impedance, phase angle and current at four frequencies.

Common mistakes

• Adding $R + X_L + X_C$. The three sides are not in a line. Use $Z = \sqrt{R^2 + (X_L - X_C)^2}$.
• Adding $X_L$ and $X_C$. They oppose each other, so take the difference.
• Adding voltmeter readings. The readings across $R$, $L$ and $C$ do not add to the source reading. Add them as arrows.
• Mixing the leader and the follower. In a coil the voltage leads. In a capacitor the current leads.
• Using $f$ where $\omega$ is needed. Reactances use $\omega = 2\pi f$ in rad/s.
• Forgetting the units. $\mu\text{F}$ is $10^{-6}$ F and mH is $10^{-3}$ H.

Remember these

• In a series circuit the same current flows in every part.
• $V_R$ is in step with the current, $V_L$ is $90^\circ$ ahead and $V_C$ is $90^\circ$ behind.
• $Z = \sqrt{R^2 + (X_L - X_C)^2}$, and $V_0 = I_0Z$, $V_{rms} = I_{rms}Z$.
• $\tan\phi = \dfrac{X_L - X_C}{R}$. Positive: inductive, the current lags. Negative: capacitive, the current leads.
• Reactances: $X_L = \omega L$ rises with frequency and $X_C = \dfrac{1}{\omega C}$ falls with frequency.

Check yourself

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

1. A series circuit has $R = 15\ \Omega$, $X_L = 48\ \Omega$ and $X_C = 28\ \Omega$. What is its impedance?

2. In a series circuit the voltmeters read $48$ V across the resistor, $90$ V across the coil and $54$ V across the capacitor. Why does the source voltmeter not read $192$ V?

3. A circuit is capacitive at $30$ Hz. Is it still capacitive at a much higher frequency?

4. In a series circuit $\phi = -35^\circ$. Does the current lead or lag the voltage? Which is larger, $X_L$ or $X_C$?