Skip to content

Paper Together: sit a paper in a room with your friends.See how it works

Series and parallel circuits – GCSE Combined Science Physics

Foundation and Higher · Grade 5

Jump to the questions and answers

Circuit symbols, charge and current, potential difference and resistance, series and parallel circuits, mains electricity, power and energy, and the National Grid.

Objectives

Specification references this topic covers: AQA 6.1.1, 6.2.1, 6.2.2, 6.2.3, 6.2.4; Edexcel P4.1, P6.6, P6.9, P8.11, P8.12, P10.2, P10.3, P10.4.

Key equations

charge flow = current × time
Q=I tQ = I\,t
  • Q charge flow (C)
  • I current (A)
  • t time (s)
potential difference = current × resistance
V=I RV = I\,R
  • V potential difference (V)
  • I current (A)
  • R resistance (Ω)
power = potential difference × current
P=V IP = V\,I
  • P power (W)
  • V potential difference (V)
  • I current (A)
energy transferred = power × time
E=P tE = P\,t
  • E energy transferred (J)
  • P power (W)
  • t time (s)
power = (current)² × resistance
P=I2 RP = I^{2}\,R
  • P power (W)
  • I current (A)
  • R resistance (Ω)
energy transferred = charge flow × potential difference
E=Q VE = Q\,V
  • E energy transferred (J)
  • Q charge flow (C)
  • V potential difference (V)
potential difference across the primary × current in the primary = potential difference across the secondary × current in the secondary
Vp Ip=Vs IsV_{p}\,I_{p} = V_{s}\,I_{s}
  • Vₚ primary potential difference (V)
  • Iₚ primary current (A)
  • Vₛ secondary potential difference (V)
  • Iₛ secondary current (A)

Series and parallel circuits explained

Start here

Look at a string of Christmas lights. On some strings, one broken bulb makes the whole string go dark. On others, one bulb breaks and all the rest keep shining, and the difference is how the bulbs are connected.

A circuit is a complete loop that charge can flow around. A flow of charge is called a current, measured in amps (A), and the cell or battery gives the charge a push. That push is called potential difference, or p.d., and it is measured in volts (V).

Picture a loop of water pipe with a pump in it. The pump is the cell and the water is the charge, and a narrow part of the pipe is a resistor, which is anything that makes it harder for charge to flow. Resistance is measured in ohms, written Ω, and a bigger resistance means a smaller current.

Circuit diagrams use standard symbols so that anyone can read them. A cell has a long line and a short line, and the long line is the positive end. A battery is two or more cells. An ammeter is a circle with an A in it, and a voltmeter is a circle with a V. A resistor is a rectangle, and a variable resistor has an arrow through it. A diode has a triangle that points the way current can flow, and an LED is a diode that gives out light. The picture shows the symbols you need.

To draw a circuit, use a ruler and draw straight wires with square corners. Join the symbols into a closed loop. To read a circuit, trace the loop with your finger from the positive end of the cell. If the wire splits into branches, you are looking at a parallel circuit.

By the end of this lesson you will be able to do four things. You will draw and read circuit diagrams, and say how current, p.d. and resistance behave in series and in parallel. You will use Q = I × t, V = I × R, P = V × I and E = P × t. You will also explain how mains electricity and the National Grid work.

cellbatteryopen switchclosed switchlampresistorvariable resistorAammeterVvoltmeterfusediodeLEDLDRthermistorMmotor
Circuit symbols: cell, battery, open switch, closed switch, lamp, resistor, variable resistor, ammeter, voltmeter, fuse, diode, LED, LDR, thermistor and motor.

Charge, current, p.d. and resistance

An electric current is a flow of electric charge. In a metal wire the charge is carried by electrons. For charge to flow around a closed circuit, the circuit must include a source of potential difference, such as a cell or a battery. A gap in the circuit, such as an open switch, stops the flow.

The size of the current is the rate of flow of charge. Charge is measured in coulombs, C. In words: charge flow = current × time. In symbols: Q = I × t. The current is in amps and the time is in seconds. A current of 2 A for 30 s moves 2 × 30 = 60 C of charge. To find the current, I = Q ÷ t.

The current has the same value at any point in a single closed loop. The charge is not used up as it goes round. It hands over energy to the components as it passes through them.

The current through a component depends on its resistance and on the p.d. across it. The greater the resistance, the smaller the current for a given p.d. In words: potential difference = current × resistance. In symbols: V = I × R, with V in volts (V), I in amps (A) and R in ohms (Ω).

You can rearrange it with a formula triangle, with V on top and I and R underneath. Cover the quantity you want. To find current, I = V ÷ R, and to find resistance, R = V ÷ I. Check the units before you start. A current in milliamps (mA) is changed to amps by dividing by 1000, so 250 mA is 0.25 A. A resistance in kilohms (kΩ) is changed to ohms by multiplying by 1000.

An ammeter measures current. It goes in series, in the loop, and it has a very low resistance so it hardly changes the current. A voltmeter measures p.d. It goes in parallel, across the component, and it has a very high resistance so almost no current goes through it. A common mistake is to put a voltmeter in the loop. That puts a huge resistance in the way and the lamp goes out.

batteryAammeterresistorlampVvoltmeter
A series circuit with an ammeter in the loop and a voltmeter connected across the resistor.

Resistors, lamps, diodes, thermistors and LDRs

For some components, the resistance stays constant. For others it changes. To find out, build a circuit with the component, an ammeter and a variable resistor in series, and a voltmeter across the component. Change the variable resistor to alter the current, and write down the current and p.d. each time. Plot current against p.d. This graph is called the I-V characteristic.

A fixed resistor at a constant temperature is called an ohmic conductor. The current is directly proportional to the p.d., so the graph is a straight line through the origin. The resistance stays constant as the current changes. A longer wire has more resistance, because the charge has more wire to push through. Double the length at constant temperature and the resistance doubles.

The resistance of a filament lamp increases as the filament gets hotter. The metal ions vibrate more and the electrons collide with them more often. The current rises more slowly than the p.d., so the graph is a curve that flattens.

The current through a diode flows in one direction only. In the reverse direction, the diode has a very high resistance and almost no current flows. An LED is a diode that gives out light.

The resistance of a thermistor decreases as the temperature increases. In a thermostat, the circuit uses a thermistor to sense the temperature, and it switches the heating on when the room is cold.

The resistance of a light-dependent resistor, or LDR, decreases as the light intensity increases. In a night light, the LDR has a high resistance in the dark, and the circuit switches the lamp on.

Series circuits

A series circuit has all its parts joined one after another in a single loop. Think of a line of people passing buckets of water from hand to hand. There is only one route for the buckets to take, so every bucket passes every person.

Because there is only one path, the current is the same at every point in the loop. In the diagram, if an ammeter reads 0.50 A at one point, an ammeter at any other point in that loop reads 0.50 A too. The charge is not used up. It hands energy to each lamp as it passes. If any part breaks, the loop is broken and the current stops everywhere. That is why one dead bulb can put out a whole old-style string of lights.

Potential difference is the energy given to each bit of charge by the cell, or handed over by the charge in a component. In a series circuit, the p.d. of the supply is shared between the components. The p.d. values across the components add up to the supply p.d.

The total resistance in a series circuit is found by adding the resistances. In symbols: R total = R1 + R2. A 2.0 Ω resistor and a 4.0 Ω resistor in series have a total resistance of 6.0 Ω. Adding another resistor in series makes the total resistance go up, because the charge has one more obstacle in the same loop. So the current gets smaller.

The bigger resistor gets the bigger share of the p.d. Take a 6.0 V battery with those two resistors. The current is 6.0 ÷ 6.0 = 1.0 A, so the 2.0 Ω resistor has 1.0 × 2.0 = 2.0 V and the 4.0 Ω resistor has 1.0 × 4.0 = 4.0 V. Twice the resistance means twice the share.

For a series calculation, follow three steps. First, add the resistors to get the total resistance. Second, use I = V ÷ R total to find the current. Third, use V = I × R on one resistor to find its share of the p.d. Then check that the shares add up to the supply. If a 12 V supply gives 7.0 V to one resistor, the other has 12 − 7.0 = 5.0 V.

batteryAlamplamp
A series circuit: a battery, an ammeter and two lamps in one loop.

Parallel circuits

In a parallel circuit each component is on its own branch, and every branch is joined across the same two points. Think of a road that splits into lanes. Cars can take any lane, but they all start and finish at the same places.

The p.d. across each branch is the same as the supply p.d. Each branch gets the full push from the cell, as if it were the only branch there. A 6.0 V battery gives 6.0 V across every branch, however many branches there are.

The current splits at a junction, which is a point where wires meet. The branch currents add up to the current from the supply. A branch with a smaller resistance takes a bigger share of the current. In the diagram, ammeter A in the main wire reads the sum of A1 and A2. If A1 reads 0.40 A and A2 reads 0.20 A, then A reads 0.60 A.

Adding a branch gives the charge another path to take. More current can flow from the supply for the same p.d., so the total resistance goes down. This is the opposite of what happens in series. Think of a shop opening a second checkout. Customers get through faster.

The total resistance of two resistors in parallel is always less than the smaller of the two. Use a 6.0 V battery with a 3.0 Ω branch and a 6.0 Ω branch. The branch currents are 6.0 ÷ 3.0 = 2.0 A and 6.0 ÷ 6.0 = 1.0 A, so the battery supplies 3.0 A. Then R total = V ÷ I = 6.0 ÷ 3.0 = 2.0 Ω, which is smaller than 3.0 Ω. You can check combinations like this in the lab. Connect resistors in series, then in parallel, measure the p.d. and the current each time, and work out R = V ÷ I.

Parallel wiring has a big advantage. Each branch works on its own, so if one branch breaks the others carry on. We come back to this when we look at houses.

batteryAlampAA1lampAA2
A parallel circuit: two lamps on separate branches, each with its own ammeter A1 and A2, and ammeter A in the main wire.

Mains electricity at home

A cell or battery gives a direct potential difference (d.c.). The charge flows one way round the circuit. Mains electricity is an alternating supply (a.c.). The p.d. keeps reversing, so the charge flows back and forth. In the UK the mains supply is about 230 V, and its frequency is 50 Hz, which means 50 complete cycles every second.

Most appliances connect to the mains with a three-core cable. Each wire has insulation with its own colour, so you can tell them apart. The live wire is brown, the neutral wire is blue and the earth wire is green and yellow stripes.

The live wire carries the alternating p.d. from the supply. The p.d. between the live wire and earth (0 V) is about 230 V. The neutral wire completes the circuit, and it is at or close to 0 V. The earth wire is a safety wire. It only carries a current if there is a fault, and it stops the appliance becoming live. If a fault makes the live wire touch a metal case, a large current flows down the earth wire. The fuse, which is a thin wire that melts when the current is too large, then cuts off the power.

A live wire can be dangerous even when a switch in the mains circuit is open. A switch in the neutral wire would leave the appliance joined to the live wire. Switches and fuses therefore go in the live wire, so that opening the switch really does disconnect the appliance from the 230 V.

Any connection between the live wire and earth is dangerous. A large current flows, which can start a fire. If you touch a live wire, the 230 V is across your body, because your feet are at about 0 V, and the current can seriously harm you. Wet hands and damaged cables make it worse.

Homes are wired in parallel. The lamps and sockets are on separate branches, so each gets the full mains p.d. of about 230 V. Each lamp can be switched on and off on its own. If one lamp breaks, the other branches are still complete circuits, so a broken light does not black out the whole house.

A circuit breaker is a switch that opens automatically when the current gets too high. It does the same job as a fuse, but it can be reset instead of being replaced. Both fuses and circuit breakers protect the wires and the appliance, and they are put in the live wire.

Power, energy and the National Grid

Power is the rate of transferring energy, so it is the energy transferred every second. The power of a circuit device depends on the p.d. across it and the current through it. In words: power = potential difference × current. In symbols: P = V × I, with power in watts (W). A kettle on the 230 V mains with a current of 8 A has a power of 230 × 8 = 1840 W.

Put V = I × R into that equation and you get a second form: power = current squared × resistance, or P = I² × R. Use it when you know the current and the resistance. A current of 3.0 A through a 10 Ω resistor gives 3.0² × 10 = 90 W. Square the current first.

The energy an appliance transfers depends on its power and on how long it is switched on. In words: energy transferred = power × time. In symbols: E = P × t, with energy in joules (J), power in watts (W) and time in seconds (s). A 2000 W kettle on for 3 minutes, which is 180 s, transfers 2000 × 180 = 360 000 J.

Work is also done when charge flows in a circuit. In words: energy transferred = charge flow × potential difference. In symbols: E = Q × V. A 12 V battery that moves 50 C of charge transfers 12 × 50 = 600 J. Together with Q = I × t, this gives E = V × I × t, which is the same as P × t.

Appliances are designed to transfer energy. A motor transfers energy to a kinetic store, as in a hairdryer fan or a washing machine. A heater transfers energy to a thermal store, as in a kettle or a toaster. The power rating shows how many joules the appliance transfers every second. A 2000 W kettle transfers 2000 J each second, and a 60 W lamp transfers only 60 J each second.

The National Grid is a system of cables and transformers that links power stations to consumers. A step-up transformer raises the p.d. from the power station to the transmission cables. A step-down transformer lowers it to a much lower value, 230 V, for homes.

The grid is an efficient way to transfer energy. For the same power, a higher p.d. means a smaller current, because P = V × I. A smaller current wastes less energy as heat in the cables, because the wasted power is I² × R. Halving the current cuts the wasted power to a quarter.

Higher tier: for a transformer, the power in equals the power out. Potential difference across the primary coil × current in the primary coil = potential difference across the secondary coil × current in the secondary coil, so Vp × Ip = Vs × Is. A transformer takes 400 V at 5.0 A into its primary coil and gives 20 000 V from its secondary coil. Is = 400 × 5.0 ÷ 20 000 = 0.10 A.

The heating effect of a current has good and bad points. It is useful in kettles, toasters and heaters, and it is how a fuse works. It is a disadvantage when it wastes energy in the wires and in other appliances, and it can make wires overheat. Using thick, low-resistance wires reduces the energy wasted.

power stationstep-uptransformerhigh p.d.cablesstep-downtransformerhomes andbusinesses
A flow of five boxes for the National Grid: power station, step-up transformer, high potential difference cables, step-down transformer, then homes and businesses.

Recap

Current is a flow of charge, and Q = I × t. V = I × R links p.d. in volts, current in amps and resistance in ohms. Ammeter in series, voltmeter in parallel. A resistor gives a straight I-V line, a filament lamp a curve, and a diode lets current through one way. A thermistor and an LDR both fall in resistance as they warm up or get brighter.

In a series circuit there is one path. The current is the same everywhere, the supply p.d. is shared, and the total resistance is R1 + R2. Add one more resistor and the current gets smaller.

In a parallel circuit there are several paths. Every branch has the full supply p.d., the branch currents add up to the supply current, and the total resistance is less than the smallest resistor. Add one more branch and the supply current gets bigger.

Mains is a.c., about 230 V and 50 Hz. The live wire is brown, the neutral wire is blue, and the earth wire is green and yellow. The earth wire is a safety wire.

P = V × I and P = I² × R. E = P × t and E = Q × V. The National Grid sends power at a high p.d. so the current and the wasted heat are small. Higher tier: Vp × Ip = Vs × Is.

How this is asked in an exam: calculation questions give you a battery p.d. and some resistors and ask for a current or a p.d. Show the three steps, add the resistors, find the current, then find the p.d., and write the unit. Power questions give two of the quantities and ask for a third. Explain questions ask you to compare series and parallel using current, p.d. and resistance, so use all three words and say what happens if one lamp breaks.

Key points

Worked examples

Example 1

A 9.0 V battery is connected in series with a 2.0 Ω resistor and a 4.0 Ω resistor. Calculate the current in the circuit.

  1. Both resistors are in the same loop, so add them: R total = 2.0 + 4.0 = 6.0 Ω
  2. Rearrange V = I × R to give I = V ÷ R, because we want the current
  3. Substitute the supply p.d. and the total resistance: I = 9.0 ÷ 6.0 = 1.5 A

Answer: 1.5 A

Common slips
  • Slip: Use one resistor only: R = 4.0 Ω The current passes through both resistors, so both resistances count.
  • Slip: Write I = R ÷ V Current is p.d. divided by resistance, not the other way round.
  • Slip: I = 9.0 × 6.0 = 54 A Multiplying p.d. by resistance does not give a current. Divide instead.

Example 2

A 6.0 V supply is connected in series with a 10 Ω resistor and a 20 Ω resistor. Calculate the potential difference across the 20 Ω resistor.

  1. Add the resistors in series: R total = 10 + 20 = 30 Ω
  2. Find the current in the loop: I = V ÷ R = 6.0 ÷ 30 = 0.20 A
  3. Find the share across the 20 Ω resistor: V = I × R = 0.20 × 20 = 4.0 V

Answer: 4.0 V

Common slips
  • Slip: R total = 20 − 10 = 10 Ω In series the resistances add. They are not subtracted.
  • Slip: I = 6.0 ÷ 20 = 0.30 A Use the total resistance with the supply p.d. to find the current in the loop.
  • Slip: V = 6.0 V across each resistor In series the p.d. is shared. Only parallel branches each have the full supply p.d.

Example 3

A 12 V battery is connected to a 6.0 Ω resistor and a 4.0 Ω resistor in parallel. Calculate the current from the battery and the total resistance of the circuit.

  1. Each branch is joined across the battery, so each has the full 12 V across it
  2. Find each branch current: I1 = 12 ÷ 6.0 = 2.0 A and I2 = 12 ÷ 4.0 = 3.0 A
  3. The branch currents add at the junction: current from the battery = 2.0 + 3.0 = 5.0 A
  4. Use the whole circuit: R total = V ÷ I = 12 ÷ 5.0 = 2.4 Ω

Answer: 5.0 A and 2.4 Ω

Common slips
  • Slip: Split the 12 V between the branches, 6 V each In parallel the p.d. across each branch equals the supply p.d.
  • Slip: I = 12 ÷ (6.0 + 4.0) = 1.2 A Adding the resistances is the series rule. Use each branch on its own.
  • Slip: Current from the battery = 3.0 A The branch currents add at the junction, so 2.0 + 3.0 = 5.0 A.
  • Slip: R total = 6.0 + 4.0 = 10 Ω The parallel total must be less than 4.0 Ω, the smaller resistance.

Questions and answers

Series and parallel circuits practice questions for GCSE Combined Science Physics. Try each one, then open the answer. Easy questions come first.

Question 1 Easy

Two resistors of 15 Ω and 25 Ω are connected in series. Calculate their total resistance.

Show answer

Answer: 40 Ω

  1. In series the resistances add: R total = R1 + R2.
  2. R total = 15 + 25 = 40 Ω.

Question 2 Easy

A 9.0 V battery is connected in series with a lamp and a resistor. The p.d. across the lamp is 5.5 V. Calculate the p.d. across the resistor.

Show answer

Answer: 3.5 V

  1. In series the p.d. values add up to the supply p.d.
  2. 9.0 − 5.5 = 3.5 V.

Question 3 Easy

In a parallel circuit the currents in the two branches are 0.35 A and 0.50 A. Calculate the current from the battery.

Show answer

Answer: 0.85 A

  1. The branch currents add up to the current from the battery.
  2. 0.35 + 0.50 = 0.85 A.

Question 4 Easy

A current of 0.20 A passes through a 30 Ω resistor. Calculate the potential difference across the resistor.

Show answer

Answer: 6.0 V

  1. Use V = I × R.
  2. V = 0.20 × 30 = 6.0 V.

Question 5 Medium

A current of 0.40 A passes through a 3.0 Ω resistor and a 5.0 Ω resistor connected in series. Calculate the p.d. of the supply.

Show answer

Answer: 3.2 V

  1. R total = 3.0 + 5.0 = 8.0 Ω.
  2. V = I × R = 0.40 × 8.0 = 3.2 V.

Question 6 Medium

A 4.5 V battery is connected in series with a 5.0 Ω resistor and a 10 Ω resistor. Calculate the current in the circuit.

Show answer

Answer: 0.30 A

  1. R total = 5.0 + 10 = 15 Ω.
  2. I = V ÷ R = 4.5 ÷ 15 = 0.30 A.

Question 7 Medium

A 3.0 Ω resistor and a 9.0 Ω resistor are connected in series with a 6.0 V battery. Calculate the p.d. across the 3.0 Ω resistor.

Show answer

Answer: 1.5 V

  1. R total = 3.0 + 9.0 = 12 Ω.
  2. I = 6.0 ÷ 12 = 0.50 A.
  3. V = I × R = 0.50 × 3.0 = 1.5 V.

Question 8 Medium

A 6.0 V battery is connected in series with resistors of 2.0 Ω, 3.0 Ω and 7.0 Ω. Calculate the current in the circuit.

Show answer

Answer: 0.50 A

  1. R total = 2.0 + 3.0 + 7.0 = 12 Ω.
  2. I = V ÷ R = 6.0 ÷ 12 = 0.50 A.

Exam-style questions

Longer questions in the style of a written paper, with the marking points. These are Yumio practice questions.

Exam-style question 14 marks

A student connects a 6.0 V battery, a 12 Ω resistor and a filament lamp in series. The ammeter reads 300 mA. (a) Calculate the potential difference across the resistor. [3 marks] (b) Calculate the potential difference across the lamp. [1 mark]

Show answer and marking points

Answer: (a) 3.6 V (b) 2.4 V

  • (a) Converts 300 mA to 0.30 A (1)
  • (a) Substitutes into V = I × R: 0.30 × 12. Allow 300 × 12 if the current was not converted. (1)
  • (a) 3.6 V. Allow 3600 V from an unconverted current. (1)
  • (b) 6.0 − their answer to (a). 2.4 V from a correct (a). (1)

Exam-style question 26 marks

A student has two identical lamps, a battery, a switch and connecting wires. Compare what happens when the two lamps are connected in series with the battery and when they are connected in parallel with the same battery. Use ideas about current, potential difference and resistance.

Show answer and marking points

Answer: In series there is one path, so the current is the same through both lamps. The battery p.d. is shared, so each lamp has half of it. The total resistance is the sum of both lamps, so the current is small and the lamps are dim. If one lamp breaks, the loop is broken and both go out. In parallel each lamp is on its own branch and has the full battery p.d., so each lamp is brighter. The total resistance is less than one lamp, so the battery supplies a larger current, which splits between the branches. If one lamp breaks the other stays lit, but the battery runs down faster.

  • Series: the current is the same through both lamps (1)
  • Series: the battery p.d. is shared between the lamps (1)
  • Parallel: each lamp has the full battery p.d. (1)
  • Parallel: lower total resistance, so a larger battery current that splits between branches (1)
  • Links p.d. to brightness: lamps in parallel are brighter than in series (1)
  • One lamp breaks: both go out in series, the other stays lit in parallel (1)

Common mistakes

  1. Adding resistances for resistors in parallel

    Add resistances only in series. A parallel total is less than the smallest resistance.

  2. Saying a lamp uses up current

    The current is the same all the way round a series loop. Energy is transferred in the lamp, not charge.

  3. Giving each series component the full supply p.d.

    In series the supply p.d. is shared. Only parallel branches each have the full p.d.

  4. Using a current in mA in V = I × R

    Change mA to A first by dividing by 1000. Then use the equation.

  5. Connecting a voltmeter in series or an ammeter in parallel

    An ammeter goes in the loop, in series. A voltmeter goes across the component, in parallel.

Questions students ask

Is Series and parallel circuits on the Foundation or Higher tier?

Series and parallel circuits is on both Foundation and Higher papers for GCSE Combined Science Physics, aimed at around grade 5.

What is a common mistake on Series and parallel circuits?

Adding resistances for resistors in parallel. Add resistances only in series. A parallel total is less than the smallest resistance.

Where are the answers to the Series and parallel circuits questions?

Open Show answer under each question on this page. The questions are Yumio practice questions, not past exam questions.

Related Physics topics

More on Series and parallel circuits

See how the papers are set: AQA Combined Science papers and Edexcel Combined Science papers.

Practise Series and parallel circuits free

Start free

Yumio is not affiliated with, endorsed by or connected to AQA, Pearson Edexcel or any other exam board.

Next topicPractise free