Branched AC circuits

Two things on one board

A lamp and a fan are plugged into the same extension board. An ammeter at the wall end of the board reads $2.0$ A. The same ammeter reads $1.6$ A when only the lamp is on, and $1.2$ A when only the fan is on. The sum of these is $2.8$ A, not $2.0$ A. Has the meter gone wrong?

No. The lamp is a resistor. For this question, the fan motor acts like a pure coil. Their currents are not in step, so they do not add as plain numbers. They add as arrows. Two arrows of $1.6$ and $1.2$ at a right angle make a third arrow of $2.0$, the long side of a 3, 4, 5 triangle.

Series circuits and parallel circuits need different pictures. Let us see why.

Three rules for branches

A branch is one path between the same two points. Take a source with several branches in parallel.

1. The voltage across every branch is the same. It is the voltage of the source.
2. At every instant, the current from the source is the sum of the currents in the branches. This is Kirchhoff's current law. It holds at every instant of an alternating current.
3. Each branch has its own impedance, so each has its own current and its own phase.

Rule 2 says the current phasors add. So we take the voltage as the reference arrow, pointing to the right, and draw each branch current against it. We use rms values for the lengths of all arrows. The picture is the same with peak values.

• Resistor branch. $I_R = V/R$, in phase with the voltage, along the reference.
• Coil branch. $I_L = V/X_L$, $90^\circ$ behind the voltage, pointing down.
• Capacitor branch. $I_C = V/X_C$, $90^\circ$ ahead of the voltage, pointing up.

In a series circuit the current is common, so the current is the reference and the voltages add. In a parallel circuit the voltage is common, so the voltage is the reference and the currents add. That is the whole difference.

A resistor and a capacitor in parallel

A worked example. A resistor of $75\ \Omega$ and a capacitor of $12\ \mu$F are joined in parallel across a source of $30$ V rms and $200$ Hz. Find the line current, its angle and the impedance of the pair.

1. The branch currents. $\omega = 2\pi f = 2\pi \times 200\ \text{Hz} = 1257\ \text{rad/s}$. The reactance is $X_C = \dfrac{1}{\omega C} = \dfrac{1}{1257\ \text{rad/s} \times 12 \times 10^{-6}\ \text{F}} = 66.3\ \Omega$. So $I_R = \dfrac{30\ \text{V}}{75\ \Omega} = 0.400$ A and $I_C = \dfrac{30\ \text{V}}{66.3\ \Omega} = 0.452$ A.
2. The line current. The two arrows are a quarter of a turn apart, so $I = \sqrt{I_R^2 + I_C^2} = \sqrt{0.400^2 + 0.452^2}\ \text{A} = 0.604$ A.
3. The angle. $\tan\varphi = \dfrac{I_C}{I_R} = \dfrac{0.452}{0.400} = 1.131$, so $\varphi = 48.5^\circ$. The line current leads the voltage.
4. The impedance of the pair. $Z = \dfrac{V}{I} = \dfrac{30\ \text{V}}{0.604\ \text{A}} = 49.7\ \Omega$. This is smaller than both $75\ \Omega$ and $66.3\ \Omega$. A second path always makes it easier for the current to flow.

The ammeters in the two branches add up to $0.852$ A. The ammeter in the line reads $0.604$ A. Both are right.

A coil and a capacitor in parallel

Now the two currents point opposite ways. The coil current is behind the voltage and the capacitor current is ahead. Along the vertical line, one is up and the other is down. So they subtract.

A worked example. A pure coil with $X_L = 40\ \Omega$ and a pure capacitor with $X_C = 100\ \Omega$ are in parallel across $80$ V rms.

1. The coil. $I_L = \dfrac{80\ \text{V}}{40\ \Omega} = 2.0$ A, behind the voltage.
2. The capacitor. $I_C = \dfrac{80\ \text{V}}{100\ \Omega} = 0.8$ A, ahead of the voltage.
3. The line. $I = I_L - I_C = 2.0\ \text{A} - 0.8\ \text{A} = 1.2$ A, behind the voltage.

The line current is smaller than the coil current alone. The capacitor takes some of the coil's current away. Now imagine that $X_C$ were also $40\ \Omega$. The two currents would be $2.0$ A each, and opposite. The line current would be zero, with $2.0$ A still flowing in each branch. We will use this in the third lesson.

Branches with two parts: the component method

A real coil has resistance. A branch can also hold a resistor in series with a coil, or with a capacitor. Each branch then has its own impedance $Z = \sqrt{R^2 + X^2}$, with $X = X_L - X_C$ for the parts in that branch. Its current is $I = V/Z$. The angle is $\tan\varphi = X/R$, behind the voltage if $X$ is positive.

To add branches at different angles, split every branch current into two parts.

• The in-phase part, $I\cos\varphi$, along the voltage.
• The quadrature part, $I\sin\varphi$, a quarter of a turn away. Count a part ahead of the voltage as positive and a part behind as negative.

Add the in-phase parts. Add the quadrature parts. The two totals are the sides of one right-angled triangle, and its long side is the line current.

A worked example. Three branches are in parallel across $100$ V rms. Branch A is a lamp of $50\ \Omega$. Branch B is a coil with a resistance of $30\ \Omega$ and a reactance of $40\ \Omega$. Branch C is a capacitor with a reactance of $80\ \Omega$.

1. Branch A. $I_A = \dfrac{100\ \text{V}}{50\ \Omega} = 2.0$ A, all in phase.
2. Branch B. $Z_B = \sqrt{30^2 + 40^2}\ \Omega = 50\ \Omega$, so $I_B = 2.0$ A. It lags by $\varphi = \arctan\dfrac{40}{30} = 53.1^\circ$. The in-phase part is $2.0\ \text{A} \times \cos 53.1^\circ = 1.2$ A. The quadrature part is $2.0\ \text{A} \times \sin 53.1^\circ = 1.6$ A, behind.
3. Branch C. $I_C = \dfrac{100\ \text{V}}{80\ \Omega} = 1.25$ A, all of it ahead.
4. The totals. In phase: $2.0\ \text{A} + 1.2\ \text{A} = 3.2$ A. Quadrature: $1.25\ \text{A} - 1.6\ \text{A} = -0.35$ A, which is $0.35$ A behind.
5. The line current. $I = \sqrt{3.2^2 + 0.35^2}\ \text{A} = 3.22$ A, behind the voltage by $\arctan\dfrac{0.35}{3.2} = 6.2^\circ$.

The three ammeters in the branches add up to $5.25$ A. The ammeter in the line reads $3.22$ A. Notice also that the capacitor has taken away most of the lag that the coil caused.

Make $X_L$ and $X_C$ equal. The orange arrow and the green arrow cancel. Which arrow is left? Now make $R$ small. How does the red arrow change?

Admittance: the parallel version of impedance

In series, the impedances combine. In parallel, it is easier to combine their inverses. The admittance of a branch is $Y = 1/Z$. Its unit is the siemens (S), and $1\ \text{S} = 1\ \Omega^{-1}$. For a parallel circuit with a resistor, an ideal coil and an ideal capacitor in separate branches:

• the conductance of the resistor is $G = \dfrac{1}{R}$,
• the susceptance of the capacitor is $B_C = \omega C = \dfrac{1}{X_C}$,
• the susceptance of the coil is $B_L = \dfrac{1}{\omega L} = \dfrac{1}{X_L}$.

The line current is $I = VY$, and the admittance of the whole circuit is

Try it on the resistor and the capacitor of the first example. $G = \dfrac{1}{75\ \Omega} = 0.01333$ S and $B_C = \dfrac{1}{66.3\ \Omega} = 0.01508$ S. So $Y = \sqrt{0.01333^2 + 0.01508^2}\ \text{S} = 0.02013$ S, and $Z = \dfrac{1}{0.02013\ \text{S}} = 49.7\ \Omega$. It is the same answer. The formula for $Y$ looks like the one for $Z$ in a series circuit. The roles of resistance and conductance are swapped.

This formula is only for branches that hold one ideal part each. For a coil with resistance, find that branch's current first, as in the component method.

Complex numbers do the arithmetic

A phasor has a length and an angle. A complex number has the same two things. So we can write a phasor as a complex number and let the rules of complex numbers add the arrows.

Engineers write the unit imaginary number as $j$, with $j^2 = -1$, because the letter $i$ is already taken by the current. A branch with a resistance $R$ and a net reactance $X = X_L - X_C$ has the impedance

Take the voltage as a real number. The branch current is $I = V/Z$. It is a complex number whose length is the rms current and whose angle is the phase. A positive $X$ gives a current with a negative angle, which means that the current lags. For branches in parallel:

Python has complex numbers built in. You write $j$ as 1j, and the cmath module finds lengths and angles.

Practise with new numbers

Try it in Python

The program takes three branches on one source: a lamp, a coil with some resistance and a capacitor. It finds each branch current as a complex number, adds them and prints the line current.

Common mistakes

• Adding the ammeter readings of the branches. The readings are lengths of arrows. Add the arrows, with their angles.
• Giving every branch the same current. That is for a series circuit. In a parallel circuit, every branch has the same voltage and its own current.
• Adding the coil and capacitor currents. They point opposite ways. Subtract the smaller from the larger.
• Combining parallel impedances as if they were resistors. $\dfrac{1}{Z} = \dfrac{1}{Z_1} + \dfrac{1}{Z_2}$ works for complex numbers, not for the plain sizes of impedances.
• Using $\sqrt{R^2 + X_C^2}$ for a resistor and a capacitor in parallel. That formula is for a series pair. For a parallel pair, use $\dfrac{1}{Z^2} = \dfrac{1}{R^2} + \dfrac{1}{X_C^2}$.
• Expecting the line current to be larger than every branch current. It can be smaller, as the coil and the capacitor show.

Remember these

• In parallel, the voltage is common and the currents add as arrows. Take the voltage as the reference.
• $I_R = \dfrac{V}{R}$ is in phase, $I_L = \dfrac{V}{X_L}$ lags by $90^\circ$ and $I_C = \dfrac{V}{X_C}$ leads by $90^\circ$.
• Component method: add the in-phase parts, add the quadrature parts, then $I = \sqrt{I_{\text{in phase}}^2 + I_{\text{quad}}^2}$.
• Admittance $Y = \dfrac{1}{Z}$ in siemens. For one ideal part in each branch, $Y = \sqrt{G^2 + (B_C - B_L)^2}$.
• With complex numbers, $Z = R + jX$, $I = V/Z$, and $I = I_1 + I_2 + \dots$

Check yourself

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

1. A resistor branch carries $6.0$ A, an ideal coil branch carries $8.0$ A and an ideal capacitor branch carries $8.0$ A, all from the same source. What does the ammeter in the line read?

2. A lamp runs on a source. A second lamp is then joined in parallel with it, across the same source. What happens to the current in the first lamp?

3. A resistor of $48\ \Omega$ and a capacitor with a reactance of $36\ \Omega$ are in parallel. What is the impedance of the pair?

4. In a series circuit we take the current as the reference arrow. In a parallel circuit we take the voltage. Why?