6.2 Series and parallel circuits — Learn

Prerequisites: Section 6.1, especially current as a flow of charge, the equations $Q = It$ and $V = IR$ , and reading a circuit diagram.

By the end of this section you can: describe the difference between series and parallel circuits; calculate currents, potential differences and resistances in series circuits; and explain why adding a resistor in series raises the total resistance while adding one in parallel lowers it.

1. Two ways to join components

Components can be joined in two ways.

In series, the components are joined one after another, so the same current flows through each one.
In parallel, the components are joined across the same two points, so the potential difference across each one is the same.

Some circuits include both series and parallel parts. A parallel pair can sit in series with another component, for example.

2. Series circuits

For components connected in series:

the current is the same through each component;
the total potential difference of the supply is shared between the components;
the total resistance of two components is the sum of their resistances.

The total resistance is called the equivalent resistance. It is the single resistance that would give the same current as the whole series combination.

R_{total} = R_1 + R_2

where $R_1$ and $R_2$ are the resistances in ohms.

Adding a resistor in series raises the total resistance. A larger resistance gives a smaller current for the same supply, so the current through every component falls.

3. Parallel circuits

For components connected in parallel:

the potential difference across each component is the same;
the total current through the whole circuit is the sum of the currents through the separate branches;
the total resistance of two resistors is less than the resistance of the smallest individual resistor.

Adding a resistor in parallel lowers the total resistance. The new branch gives the current another path to follow, so the total current from the supply rises. A larger total current for the same supply means a smaller total resistance.

4. Worked examples

Worked example 1. A 4.0 $\Omega$ resistor and a 6.0 $\Omega$ resistor are connected in series with a 5.0 V supply. Calculate the total resistance, the current, and the potential difference across each resistor.

\begin{aligned} R_{total} &= R_1 + R_2 \\ &= 4.0 + 6.0 \\ &= 10\,\Omega \end{aligned}

\begin{aligned} I &= \frac{V}{R_{total}} \\ &= \frac{5.0}{10} \\ &= 0.50\,\text{A} \end{aligned}

The current is the same through both resistors, so each one carries 0.50 A.

\begin{aligned} V_1 &= IR_1 \\ &= 0.50 \times 4.0 \\ &= 2.0\,\text{V} \end{aligned}

\begin{aligned} V_2 &= IR_2 \\ &= 0.50 \times 6.0 \\ &= 3.0\,\text{V} \end{aligned}

Check: $2.0\,\text{V} + 3.0\,\text{V} = 5.0\,\text{V}$ , which is the supply potential difference. The supply is shared between the two resistors.

Worked example 2. A 10 $\Omega$ resistor and a 20 $\Omega$ resistor are connected in parallel across the same 5.0 V supply. Find the current through each resistor, and the total current from the supply.

The potential difference across each resistor is 5.0 V, because they are in parallel.

\begin{aligned} I_1 &= \frac{V}{R_1} \\ &= \frac{5.0}{10} \\ &= 0.50\,\text{A} \end{aligned}

\begin{aligned} I_2 &= \frac{V}{R_2} \\ &= \frac{5.0}{20} \\ &= 0.25\,\text{A} \end{aligned}

\begin{aligned} I_{total} &= I_1 + I_2 \\ &= 0.50 + 0.25 \\ &= 0.75\,\text{A} \end{aligned}

The total resistance of the parallel pair is less than 10 $\Omega$ , the smaller of the two resistors, because the total current from the supply is larger than either branch current.

5. Comparing series and parallel

The figure shows two lamps connected to the same battery. You can join them in series or in parallel, and you can remove one lamp from its holder.

Predict first. Do not press the buttons yet. Write down your answers.

In series, what happens to the other lamp when one lamp is removed?
In parallel, what happens to the other lamp when one lamp is removed?
In which arrangement are the two lamps brighter?

Test. Press each button and watch the lamps and the readouts. Then remove one lamp in each arrangement and check your predictions.

Explain. In the parallel arrangement, each lamp still gets the full battery potential difference. Why does removing one lamp in series switch the other lamp off, while removing one lamp in parallel leaves the other lamp lit? Write your answer in two or three sentences.

Watch out. Two common errors are worth checking now. First, in parallel the branch currents need not be equal; it is the potential differences that are equal. Second, the total resistance of a parallel pair is less than the smallest resistor, not more. If you find the resistor sum rules confusing, remember what each connection does to the current: series adds resistance, parallel opens another path. Re-check your answer by asking whether the total current should be bigger or smaller.

6. Progress check

Return to the chapter problem. The pump circuit has a 12 V battery and two resistors of 6.0 $\Omega$ and 2.0 $\Omega$ in series.

Calculate the total resistance of the circuit.
Calculate the current through the pump circuit.
Calculate the potential difference across each resistor.

You should now be able to

describe the difference between a series circuit and a parallel circuit;
state that the current is the same through components in series;
state that the potential difference is the same across components in parallel;
use $R_{total} = R_1 + R_2$ for resistors in series;
calculate the current and the potential differences in a series circuit;
calculate the branch currents and the total current in a parallel circuit;
explain why a series connection raises the total resistance and a parallel connection lowers it.