Resistors in series

One path for the current

A string of small lights on a festival day goes out completely when one bulb is taken out. All the bulbs are in one line, one after another. This is a series connection.

Resistors in series

Resistors are in series when they are joined end to end, so that the current has only one path through them.

1. The current is the same. The charge has nowhere else to go, so the same current $I$ passes through $R_1$, $R_2$ and $R_3$. An ammeter placed anywhere in the circuit shows the same reading.

2. The potential differences add up. The potential difference of the battery is shared by the resistors:

3. The combined resistance is the sum. With $V_1 = IR_1$, $V_2 = IR_2$ and $V_3 = IR_3$, we get $V = I(R_1 + R_2 + R_3)$. So the one resistor that could take the place of all three is

The combined resistance is larger than the largest of the resistors. When you add a resistor in series, the current falls.

You should find that the three shares always stack up to the battery's voltage, and the current is the same whichever resistor you look at.

A worked example

Three resistors of $2\ \Omega$, $3\ \Omega$ and $5\ \Omega$ are joined in series to a $10\ \text{V}$ battery. Find the current, and the potential difference across each resistor.

Step 1. $R_s = 2\ \Omega + 3\ \Omega + 5\ \Omega = 10\ \Omega$.

Step 2. We want $I$, so $I = \dfrac{V}{R_s} = \dfrac{10\ \text{V}}{10\ \Omega} = 1\ \text{A}$. This is the current in each resistor.

Step 3. $V_1 = 1\ \text{A} \times 2\ \Omega = 2\ \text{V}$, $V_2 = 1\ \text{A} \times 3\ \Omega = 3\ \text{V}$, $V_3 = 1\ \text{A} \times 5\ \Omega = 5\ \text{V}$.

Check. $2\ \text{V} + 3\ \text{V} + 5\ \text{V} = 10\ \text{V}$, the potential difference of the battery. The largest resistor takes the largest share.

The disadvantage of a series circuit

If one device in a series circuit stops working, or if the circuit breaks at one place, the current stops in all the devices. Also, all devices get the same current, though they may need different currents. So the lights of a house are not joined in series.

Practise with new numbers

Try it in Python

The program joins three resistors in series, and it finds the current and the potential difference across each.

Remember these

• In series: the current is the same, the potential differences add up, and $R_s = R_1 + R_2 + R_3$.
• The combined resistance is larger than the largest resistor.
• If one device in a series circuit fails, they all stop.

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 $4\ \Omega$ and $6\ \Omega$ are in series. What is the combined resistance?

2. What current flows if they are joined to a $5\ \text{V}$ battery?

3. A $2\ \Omega$ and an $8\ \Omega$ resistor in series share $5\ \text{V}$. Which gets the larger share, and how much?

4. Why does adding another resistor in series make the current fall?