Resistors in parallel

Many paths for the current

In a house, you can switch off one light, and the others stay on. Each light has its own path to the supply. This is a parallel connection.

Resistors in parallel

Resistors are in parallel when they are all joined between the same two points, so that the current can go along any one of the branches.

1. The potential difference is the same. The two ends of each resistor are joined to the same two points, so each has the same potential difference $V$.

2. The currents add up. The current of the battery divides among the branches:

3. The combined resistance. With $I_1 = \dfrac{V}{R_1}$, $I_2 = \dfrac{V}{R_2}$ and $I_3 = \dfrac{V}{R_3}$, we get $I = V\left(\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}\right)$. So

The combined resistance is smaller than the smallest of the resistors. When you add a resistor in parallel, you give the current one more path, so the total current goes up.

For two resistors, we can write $R_p = \dfrac{R_1 R_2}{R_1 + R_2}$ (the product over the sum).

You should find that the smaller resistor always carries the larger current, and that a third branch always raises the total current.

A worked example

A $10\ \Omega$ resistor and a $40\ \Omega$ resistor are joined in parallel to a $24\ \text{V}$ battery. Find the combined resistance, the current in each resistor, and the total current.

Step 1. $\dfrac{1}{R_p} = \dfrac{1}{10\ \Omega} + \dfrac{1}{40\ \Omega} = \dfrac{4 + 1}{40\ \Omega} = \dfrac{1}{8\ \Omega}$, so $R_p = 8\ \Omega$.

Step 2. Each branch has $24\ \text{V}$ across it. $I_1 = \dfrac{24\ \text{V}}{10\ \Omega} = 2.4\ \text{A}$ and $I_2 = \dfrac{24\ \text{V}}{40\ \Omega} = 0.6\ \text{A}$.

Step 3. The total current is $2.4\ \text{A} + 0.6\ \text{A} = 3\ \text{A}$. We can check it: $\dfrac{V}{R_p} = \dfrac{24\ \text{V}}{8\ \Omega} = 3\ \text{A}$.

Check. $R_p = 8\ \Omega$ is less than the smallest resistor, $10\ \Omega$. The smaller resistor carries the larger current.

Why are the lights of a house in parallel?

• Each appliance gets the full potential difference of the supply, $220\ \text{V}$, whatever the other appliances do.
• Each appliance can be switched on and off by itself. If one bulb fails, the others stay on.
• Each appliance can have its own resistance and so its own current.

A mixed circuit

Many circuits have both kinds of connection. To solve one, we first find the resistance of each parallel part, then add the series parts. We take one step at a time.

Practise with new numbers

Try it in Python

The program joins a $10\ \Omega$ and a $40\ \Omega$ resistor in parallel, and it finds the combined resistance and the currents.

Remember these

• In parallel: the potential difference is the same, the currents add up, and $\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots$
• The combined resistance is smaller than the smallest resistor.
• The lights and the appliances of a house are in parallel.

Check yourself

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

1. Two resistors of $4\ \Omega$ each are in parallel. What is the combined resistance?

2. Why does the combined resistance fall when you add one more resistor in parallel?

3. One bulb of a house fuses. Do the others go out? Why?

4. An $8\ \Omega$ and a $24\ \Omega$ resistor are in parallel across $6\ \text{V}$. What is the total current?