📈 Larger resistance
Takes the larger share of the supply voltage, because V = I × R and the current is the same.
Energy and Voltage in Circuits » Series Circuit Calculations
What you'll learn this session
Study time: 30 minutes
Edexcel spec: 2.19
- The three rules for two resistive components connected in series
- How to calculate the current, the voltages and the resistances step by step
- Why the larger resistance takes the larger share of the voltage
- How to set out your working so you never mix up the numbers
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Unlock This CourseThree rules for two components in series
Picture the circuit this lesson is about. Draw a battery. From its positive terminal, draw one wire to resistor R1, then on to resistor R2, then back to the negative terminal. There is one loop and there are no junctions. An ammeter can go anywhere in the loop. A voltmeter is connected across each resistor, and a third voltmeter is connected across the battery.
Every calculation on this circuit uses three rules.
| Quantity | Rule in a series circuit | In symbols |
|---|---|---|
| Current | The same current flows through both components | I is the same everywhere |
| Voltage | The supply voltage equals the sum of the voltages across the two components | Vsupply = V1 + V2 |
| Resistance | The total resistance equals the sum of the two resistances | Rtotal = R1 + R2 |
You met the current rule in Current in a Series Circuit: there is only one path, so the current has nowhere else to go. The voltage rule and the resistance rule are the new tools. Notice the pattern: in series the current stays the same, but voltages add and resistances add.
The resistance rule makes sense. The current has to pass through R1 and then through R2, so the two resistances work together to oppose it. Two resistors in series always have a bigger total resistance than either one alone.
Key terms:
- Resistive component: any component that has resistance, such as a resistor, a lamp or a heater.
- Total resistance: the resistance of the whole circuit. It is the one resistance that would let the same current flow from the same supply.
Using V = I × R in the right place
Remember, voltage = current × resistance. In a series circuit you can use this equation in three places, as long as the three numbers belong together:
1️⃣ Component 1
V1 = I × R1
The voltage across R1, the current in it and its own resistance.
2️⃣ Component 2
V2 = I × R2
The voltage across R2, the current in it and its own resistance.
🔋 Whole circuit
Vsupply = I × Rtotal
The supply voltage, the current and the total resistance.
The golden rule is never mix the columns. Do not divide the supply voltage by the resistance of one component. A good way to stay safe is to draw a small grid before you start. Fill in what the question tells you, then find each unknown in turn.
| Component 1 | Component 2 | Whole circuit | |
|---|---|---|---|
| Voltage (V) | V1 | V2 | add across: V1 + V2 |
| Current (A) | I | I | same across: I |
| Resistance (Ω) | R1 | R2 | add across: R1 + R2 |
Going across a row, use the series rules. Going down a column, use V = I × R. As soon as a column has two numbers in it, you can work out the third.
Worked examples
Worked example 1: finding the current and both voltages
An 8 Ω resistor and a 16 Ω resistor are connected in series with a 12 V battery. Calculate the current in the circuit and the voltage across each resistor.
Step 1, total resistance: Rtotal = R1 + R2 = 8 + 16 = 24 Ω
Step 2, current (whole circuit column): I = Vsupply ÷ Rtotal = 12 ÷ 24 = 0.5 A
Step 3, the current is the same in both resistors, so use 0.5 A for each one.
V1 = I × R1 = 0.5 × 8 = 4 V
V2 = I × R2 = 0.5 × 16 = 8 V
Check: 4 V + 8 V = 12 V, which is the supply voltage.
Worked example 2: finding an unknown resistance
A 20 Ω resistor and a lamp are connected in series with a 6 V battery. The ammeter reads 0.2 A. Calculate the resistance of the lamp.
Step 1, voltage across the resistor: V1 = I × R1 = 0.2 × 20 = 4 V
Step 2, voltage across the lamp: V2 = Vsupply − V1 = 6 − 4 = 2 V
Step 3, resistance of the lamp: R2 = V2 ÷ I = 2 ÷ 0.2 = 10 Ω
Another way: Rtotal = Vsupply ÷ I = 6 ÷ 0.2 = 30 Ω, so R2 = 30 − 20 = 10 Ω. Both methods give the same answer, and either one earns full marks.
Worked example 3: finding the supply voltage
A 15 Ω resistor and a 45 Ω resistor are connected in series with a power supply. A voltmeter across the 15 Ω resistor reads 3 V. Calculate the supply voltage.
Step 1, current in the 15 Ω resistor: I = V1 ÷ R1 = 3 ÷ 15 = 0.2 A
Step 2, the same 0.2 A flows in the 45 Ω resistor: V2 = I × R2 = 0.2 × 45 = 9 V
Step 3, supply voltage: Vsupply = V1 + V2 = 3 + 9 = 12 V
Check with the whole circuit: Rtotal = 15 + 45 = 60 Ω and 0.2 × 60 = 12 V.
Who gets the bigger share of the voltage?
Look back at the answers. In example 1 the 16 Ω resistor had 8 V across it and the 8 Ω resistor had only 4 V. In example 3 the 45 Ω resistor had 9 V and the 15 Ω resistor had 3 V. Each time:
The component with the larger resistance takes the larger share of the supply voltage.
The reason is V = I × R. Both components carry the same current, so the only thing that differs is the resistance. A bigger R multiplied by the same I gives a bigger V.
In fact the voltage is shared in the same ratio as the resistances. In example 3, 45 Ω is three times 15 Ω, and 9 V is three times 3 V. This gives you some quick checks:
- Two equal resistances share the supply voltage equally. Two 10 Ω resistors on a 6 V battery have 3 V each.
- If one resistance is double the other, it has double the voltage.
- If your answer gives the smaller resistance the bigger voltage, you have made a slip somewhere.
📉 Smaller resistance
Takes the smaller share. The two shares always add up to the supply voltage.
Common mistakes
- Mixing the columns. Dividing the 12 V supply voltage by 8 Ω in example 1 gives 1.5 A, which is wrong. The supply voltage goes with the total resistance.
- Using a different current for each component. In series there is one current. Find it once and use it everywhere.
- Giving each component the full supply voltage. That is the rule for components in parallel. In series the supply voltage is shared.
- Halving the voltage without thinking. The voltage splits equally only when the two resistances are equal.
- Forgetting units. Change mA to A and kΩ to Ω before you calculate, and write A, V or Ω on every answer.
- Not checking. Add your two voltages. If they do not make the supply voltage, look again.
Exam-style question
A student connects resistor X and resistor Y in series with a 9 V battery. Resistor X has a resistance of 120 Ω. A voltmeter connected across resistor X reads 3.6 V.
(a) Calculate the current in resistor X. [3 marks]
(b) State the current in resistor Y. [1 mark]
(c) Calculate the voltage across resistor Y. [2 marks]
(d) Calculate the resistance of resistor Y. [2 marks]
(e) Resistor Y is replaced with a resistor of larger resistance. Explain what happens to the reading on the voltmeter across resistor X. [3 marks]
Model answer
(a) I = V ÷ R (1) = 3.6 ÷ 120 (1) = 0.03 A (1)
(b) 0.03 A (1)
(c) voltage across Y = supply voltage − voltage across X = 9 − 3.6 (1) = 5.4 V (1)
(d) R = V ÷ I = 5.4 ÷ 0.03 (1) = 180 Ω (1)
(e) The total resistance of the circuit increases (1), so the current in the circuit decreases (1). The voltage across X = current × resistance of X, and the resistance of X has not changed, so the voltmeter reading decreases (1).
Exam tip
In part (d), use resistor Y's own voltage, 5.4 V, not the 9 V of the battery. Then check your answer: 120 Ω + 180 Ω = 300 Ω, and 9 V ÷ 300 Ω = 0.03 A, which matches part (a).