Chapter 5 of 8

All 8 chapters
  1. 1Charge and current, readunderstood
  2. 2Potential difference, readunderstood
  3. 3Ohm's law, readunderstood
  4. 4Different components, readunderstood
  5. 5Series circuits, readunderstood
  6. 6Parallel circuits, readunderstood
  7. 7Power in circuits, readunderstood
  8. 8Mains electricity and safety, readunderstood

S4.12.5 Chapter 5

Series circuits

Tariq puts a second lamp in the same loop, and both lamps glow more dimly than one did. Why?

Your guess

Pick one before you read on. It is fine to be wrong.

Your guess

  • You picked A. Read on. At the end, the chapter comes back to your guess.

  • You picked B. Read on. At the end, the chapter comes back to your guess.

  • You picked C. Read on. At the end, the chapter comes back to your guess.

  • You picked D. Read on. At the end, the chapter comes back to your guess.

By the end you can

  • Calculate total resistance in series
  • Explain how voltage is shared in a series circuit

In a series circuit the parts are joined one after another in a single loop. The charge has only one path to take. Three rules follow from that, and each one comes from something you already know.

Rule 1: the current is the same everywhere in the loop. Charge cannot pile up or vanish, so the same charge passes every point each second. You saw this with the ammeters in the first chapter.

Rule 2: the resistances add up. The charge goes through each resistor in turn, and each one resists. Two resistors end to end are like one longer one. The total resistance is R = R₁ + R₂, and for three resistors R = R₁ + R₂ + R₃.

Adding a resistor in series always raises the total resistance. With the same battery, the current then falls, because I = V ÷ R. Watch it on one battery of 6 V.

Draw it first

More resistors in series, less current

  1. A loop with a battery labelled 6 V, drawn on the left with + at the top and − at the bottom. An ammeter is on the right wire with its reading beside it. Step 1: one resistor labelled 4 Ω on the top wire. The ammeter reads 1.5 A. Step 2: a second resistor labelled 6 Ω is added beside the first. The first has not moved. The ammeter reads 0.6 A. Step 3: a third resistor labelled 10 Ω is added on the bottom wire. The others have not moved. The ammeter reads 0.3 A. Arrowheads on the wires point clockwise. Labels: 6 V, +, −, 4 Ω, 1.5 A, A, 0.6 A, 6 Ω, 0.3 A, 10 Ω.+−6 V4 ΩA1.5 A

    1 One resistor of 4 Ω. The total resistance is 4 Ω, so I = 6 V ÷ 4 Ω = 1.5 A.

  2. A loop with a battery labelled 6 V, drawn on the left with + at the top and − at the bottom. An ammeter is on the right wire with its reading beside it. Step 1: one resistor labelled 4 Ω on the top wire. The ammeter reads 1.5 A. Step 2: a second resistor labelled 6 Ω is added beside the first. The first has not moved. The ammeter reads 0.6 A. Step 3: a third resistor labelled 10 Ω is added on the bottom wire. The others have not moved. The ammeter reads 0.3 A. Arrowheads on the wires point clockwise. Labels: 6 V, +, −, 4 Ω, 1.5 A, A, 0.6 A, 6 Ω, 0.3 A, 10 Ω.+−6 V4 Ω6 ΩA0.6 A

    2 Add a 6 Ω resistor in series. The total is 4 Ω + 6 Ω = 10 Ω, so I = 6 V ÷ 10 Ω = 0.6 A.

  3. A loop with a battery labelled 6 V, drawn on the left with + at the top and − at the bottom. An ammeter is on the right wire with its reading beside it. Step 1: one resistor labelled 4 Ω on the top wire. The ammeter reads 1.5 A. Step 2: a second resistor labelled 6 Ω is added beside the first. The first has not moved. The ammeter reads 0.6 A. Step 3: a third resistor labelled 10 Ω is added on the bottom wire. The others have not moved. The ammeter reads 0.3 A. Arrowheads on the wires point clockwise. Labels: 6 V, +, −, 4 Ω, 1.5 A, A, 0.6 A, 6 Ω, 0.3 A, 10 Ω.+−6 V4 Ω6 Ω10 ΩA0.3 A

    3 Add a 10 Ω resistor. The total is 20 Ω, so I = 6 V ÷ 20 Ω = 0.3 A. Each resistor made the current smaller.

Lamps are resistors that glow. Two lamps in series have about twice the resistance of one. It is a little less, because each dimmer filament is cooler. So the current falls. Each lamp now has less current, and it also has only part of the battery p.d. Both glow less. Look at one lamp and two lamps on a 3 V battery.

Two loops, one above the other, each with a battery of two cells. The top loop has one lamp, which is lit with long rays round it. The bottom loop has two lamps in a line, and each is lit with shorter rays round it. Labels: none.
One lamp, then two lamps in series on the same battery. The rays show how brightly each lamp glows. Each of two lamps glows less than one lamp alone.

Rule 3: the p.d.s across the parts add up to the p.d. of the supply. Each coulomb gains its energy from the battery, and it gives that energy away in steps, a little in each resistor. So V = V₁ + V₂. Look at the 6 V battery with a 4 Ω and a 6 Ω resistor.

A loop with a battery labelled 6 V, an ammeter that reads 0.6 A and two resistors in series, labelled 4 Ω and 6 Ω. A voltmeter is joined across each resistor. The voltmeter across the 4 Ω resistor reads 2.4 V and the one across the 6 Ω resistor reads 3.6 V. Labels: 6 V, A, 0.6 A, 4 Ω, V, 2.4 V, 6 Ω, V, 3.6 V.6 VA0.6 AV3.6 V6 ΩV2.4 V4 Ω
The current is 0.6 A everywhere. Each resistor has its label under its voltmeter reading. The p.d.s are 2.4 V across the 4 Ω resistor and 3.6 V across the 6 Ω one. Together they make the 6 V of the battery.

See the same thing as a stack. The bar is as high as the battery p.d. It is cut into the p.d. across each resistor. The bigger resistor has the bigger part of the bar.

A tall bar, as high as the 6 V of the battery, cut into two parts. The lower part, dotted, is labelled 4 Ω: 2.4 V. The upper part, hatched, is higher and is labelled 6 Ω: 3.6 V. A brace on the left, labelled supply 6 V, spans the whole bar. Labels: 4 Ω: 2.4 V, 6 Ω: 3.6 V, supply, 6 V.4 Ω: 2.4 V6 Ω: 3.6 Vsupply6 V
The p.d. of the battery is shared between the resistors. The 6 Ω resistor has 1.5 times the resistance of the 4 Ω resistor, and 1.5 times the p.d.

In a series circuit the p.d. is shared in proportion to the resistance. A resistor with twice the resistance of the other takes twice the p.d. This follows from V = IR, because the current I is the same for both.

See it three ways

As a picture

A table with the columns Quantity and Rule in series. Row 1: current, same everywhere. Row 2: p.d., adds to supply. Row 3: resistance, adds up.QuantityIn seriescurrentsame everywherep.d.adds to supplyresistanceadds up

In words

The same current goes through every part. The p.d.s of the parts add up to the p.d. of the battery. The resistances add up to the total resistance.

In symbols

I = I₁ = I₂. V = V₁ + V₂. R = R₁ + R₂.

Worked example

Two resistors of 15 Ω and 25 Ω are joined in series. What is the total resistance?

  1. Write the rule: R = R₁ + R₂.
  2. Put in the numbers: R = 15 Ω + 25 Ω.
  3. R = 40 Ω.

Answer: 40 Ω

Worked example

A 12 V battery is joined to a 20 Ω resistor and a 40 Ω resistor in series. What current flows?

  1. Find the total resistance: R = 20 Ω + 40 Ω = 60 Ω.
  2. Turn V = IR round to I = V ÷ R.
  3. Put in the numbers: I = 12 V ÷ 60 Ω = 0.2 A.

Answer: 0.2 A

A loop with a battery labelled 6 V and two resistors in series, labelled 10 Ω and 20 Ω. No meters are drawn. Labels: 6 V, 10 Ω, 20 Ω.6 V10 Ω20 Ω
The loop of the next worked example, a 6 V battery with a 10 Ω and a 20 Ω resistor.

Worked example

A 6 V battery is joined to a 10 Ω resistor and a 20 Ω resistor in series. Find the p.d. across each resistor.

  1. Find the total resistance: R = 10 Ω + 20 Ω = 30 Ω.
  2. Find the current: I = V ÷ R = 6 V ÷ 30 Ω = 0.2 A.
  3. The p.d. across the 10 Ω resistor: V = IR = 0.2 A × 10 Ω = 2 V.
  4. The p.d. across the 20 Ω resistor: V = IR = 0.2 A × 20 Ω = 4 V.
  5. Check: 2 V + 4 V = 6 V, which is the p.d. of the battery.

Answer: 2 V and 4 V

A series circuit has a weak point. If any one part is taken out, or a lamp burns out, the loop is broken and the current stops everywhere. This is why a switch goes in series with what it controls.

Try it

One bulb, two bulbs, one removed

You need: two torch cells of 1.5 V each, two small torch bulbs of the same kind, bare wire or kitchen foil strips, sticky tape.

Be safe: Use only torch cells. Do not leave the loop joined for long, because the wires and the bulb can get warm. Lift a wire to switch off.

  1. Put the two cells end to end, + touching −. Join one bulb across the two ends, with wire or foil. Look at how bright it is.
  2. Now put the second bulb in the same loop, one after the other, with a wire between them. Look at both bulbs.
  3. Lift one bulb out of its place, so the loop is broken. Look at the other bulb.

What you will see: One bulb on its own is bright. With two bulbs in the same loop, each is dimmer, because they share the p.d. and the current falls. Take one away and the other goes out too, because the loop is broken.

If you cannot try it, this is what happens.
A table with the columns Loop and Bulbs. Row 1: one bulb, bright. Row 2: two bulbs, both dim. Row 3: one lifted out, both off.LoopBulbsone bulbbrighttwo bulbsboth dimone liftedboth off

A wrong idea many people have

In a series circuit the current is shared between the resistors, just like the p.d.

Why it feels right: We say the p.d. is shared, and we tend to think that everything is shared.

The test that breaks it: Remember the two ammeters in chapter 1, or look at the main loop. The ammeter reads 0.6 A. The same charge must go through both resistors, one after the other.

What is true: The current is the same in every part of a series loop. It is the p.d. that is shared. The current is not used up and not split.

A wrong idea many people have

Each resistor in series has the whole battery p.d. across it.

Why it feels right: One resistor alone on the battery does have the whole p.d., so it seems that every resistor must.

The test that breaks it: Look at the main loop. The battery is 6 V, but the voltmeters read 2.4 V and 3.6 V. If each had 6 V, the charge would give away 12 J for each coulomb, and it only gained 6 J.

What is true: The p.d.s across the resistors in series add up to the battery p.d. Each coulomb gives away its energy in steps.

A wrong idea many people have

Adding a resistor in series makes it easier for the current to flow, so the current grows.

Why it feels right: More parts seem to mean more paths. Parallel circuits do give more paths.

The test that breaks it: Look at the strip. The current goes 1.5 A, then 0.6 A, then 0.3 A as resistors are added in series. A resistor in series makes the total resistance bigger, so the current falls.

What is true: In series, the resistances add up, so the total is bigger than any one resistor. The current is smaller. More paths come only with a parallel circuit.

Key words

series circuit
a circuit with the parts in one loop, one after another, so the current has only one path.
total resistance
the one resistance that would give the same current as all the resistors together. In series, R = R₁ + R₂ + R₃.

Back to your guess

The answer is D: The lamps share the battery p.d. and add their resistances, so the current falls.

In a series loop the resistances add, so the current is smaller. The p.d. of the battery is shared between the lamps, so each gets only part of it. Each lamp has less current and less p.d., so it glows less.

Check yourself

Try each one before you look. A wrong answer tells you where to go next.

  1. Resistors of 7 Ω and 11 Ω are joined in series. What is the total resistance?

    Tap your answer.

    1. Not this one. You divided 11 by 7. In series, add the resistances. Try another.

    2. Not this one. You multiplied the two resistances. In series the resistances add. Try another.

    3. Yes, that is right. In series the resistances add: R = 7 Ω + 11 Ω = 18 Ω.

    4. Not this one. You took the average. The total is the sum, not the middle. Try another.

    See the answer

    18 Ω

    In series the resistances add: R = 7 Ω + 11 Ω = 18 Ω.

    • If you chose A: You divided 11 by 7. In series, add the resistances.
    • If you chose B: You multiplied the two resistances. In series the resistances add.
    • If you chose D: You took the average. The total is the sum, not the middle.

  2. A 12 V battery is joined to a 30 Ω resistor and a 90 Ω resistor in series. What current flows?

    Tap your answer.

    1. Yes, that is right. The total resistance is 30 Ω + 90 Ω = 120 Ω. I = V ÷ R = 12 V ÷ 120 Ω = 0.1 A.

    2. Not this one. You used only the 30 Ω resistor. Both resistors are in the loop, so use their total. Go to S3.10.3. Try another.

    3. Not this one. You divided the total resistance by the p.d. Turn V = IR round to I = V ÷ R. Try another.

    4. Not this one. You used only the 90 Ω resistor. Add the two resistances first. Go to S3.10.3. Try another.

    See the answer

    0.1 A

    The total resistance is 30 Ω + 90 Ω = 120 Ω. I = V ÷ R = 12 V ÷ 120 Ω = 0.1 A.

    • If you chose B: You used only the 30 Ω resistor. Both resistors are in the loop, so use their total. Go to S3.10.3.
    • If you chose C: You divided the total resistance by the p.d. Turn V = IR round to I = V ÷ R.
    • If you chose D: You used only the 90 Ω resistor. Add the two resistances first. Go to S3.10.3.

  3. The picture shows a 12 V battery with a 25 Ω and a 75 Ω resistor in series. A voltmeter is across the 75 Ω resistor. What does it read?

    A loop with a battery labelled 12 V, an ammeter that reads 0.12 A and two resistors in series, labelled 25 Ω and 75 Ω. A voltmeter is joined across the resistor labelled 75 Ω, which has its label under the voltmeter. Labels: 12 V, A, 0.12 A, 25 Ω, 75 Ω, V.12 VA0.12 AV75 Ω25 Ω
    Tap your answer.

    1. Not this one. That is the p.d. of the battery. The 75 Ω resistor has only part of it, because the p.d. is shared. Go to S3.10.3. Try another.

    2. Not this one. That would be half of the battery p.d. The resistors are not equal, so the sharing is not half and half. Try another.

    3. Not this one. That is the p.d. across the 25 Ω resistor. The voltmeter is across the 75 Ω one. Try another.

    4. Yes, that is right. The ammeter reads 0.12 A, the same all round the loop. V = IR = 0.12 A × 75 Ω = 9 V. The other resistor has 0.12 A × 25 Ω = 3 V, and 9 V + 3 V = 12 V.

    See the answer

    9 V

    The ammeter reads 0.12 A, the same all round the loop. V = IR = 0.12 A × 75 Ω = 9 V. The other resistor has 0.12 A × 25 Ω = 3 V, and 9 V + 3 V = 12 V.

    • If you chose A: That is the p.d. of the battery. The 75 Ω resistor has only part of it, because the p.d. is shared. Go to S3.10.3.
    • If you chose B: That would be half of the battery p.d. The resistors are not equal, so the sharing is not half and half.
    • If you chose C: That is the p.d. across the 25 Ω resistor. The voltmeter is across the 75 Ω one.

  4. Two identical lamps are joined in series to a battery. One lamp is lifted out of its holder. What happens to the other lamp?

    Tap your answer.

    1. Not this one. With one lamp out there is a gap in the loop. A gap lets no current through, so the other lamp cannot glow. Go to S3.10.3. Try another.

    2. Yes, that is right. A series circuit has only one path. Remove one part and the loop is broken, so the current stops in every part.

    3. Not this one. The other lamp depends on the current of the loop. With a gap, no current flows at all. Try another.

    4. Not this one. The current stops completely when the loop is broken. A circuit in series has only one path. Try another.

    See the answer

    It goes out, because the loop is broken

    A series circuit has only one path. Remove one part and the loop is broken, so the current stops in every part.

    • If you chose A: With one lamp out there is a gap in the loop. A gap lets no current through, so the other lamp cannot glow. Go to S3.10.3.
    • If you chose C: The other lamp depends on the current of the loop. With a gap, no current flows at all.
    • If you chose D: The current stops completely when the loop is broken. A circuit in series has only one path.

  5. Which quantity is the same for every part in a series circuit?

    Tap your answer.

    1. Not this one. The p.d.s add up to the battery p.d., so they are shared, not equal. Try another.

    2. Yes, that is right. There is only one path, so the same charge passes every point each second. The current is the same in every part.

    3. Not this one. The parts can have different resistances. The total is their sum. Try another.

    4. Not this one. Power depends on the current and the p.d., and the p.d.s can differ, so the powers can differ. Try another.

    See the answer

    The current

    There is only one path, so the same charge passes every point each second. The current is the same in every part.

    • If you chose A: The p.d.s add up to the battery p.d., so they are shared, not equal.
    • If you chose C: The parts can have different resistances. The total is their sum.
    • If you chose D: Power depends on the current and the p.d., and the p.d.s can differ, so the powers can differ.

  6. A 10 Ω resistor and a 40 Ω resistor are joined in series. Which one has the bigger share of the battery p.d.?

    Tap your answer.

    1. Not this one. The current is the same in both. With V = IR, the bigger resistance has the bigger p.d. Try another.

    2. Not this one. With the same current, V = IR gives more p.d. across the larger resistance, not the same. Try another.

    3. Yes, that is right. The current is the same, so V = IR gives a p.d. in proportion to R. 40 Ω is four times 10 Ω, so it has four times the p.d.

    4. Not this one. The sharing depends on the resistances alone. The battery p.d. only sets how big the shares are. Try another.

    See the answer

    The 40 Ω resistor: 4 times the resistance gives 4 times the p.d.

    The current is the same, so V = IR gives a p.d. in proportion to R. 40 Ω is four times 10 Ω, so it has four times the p.d.

    • If you chose A: The current is the same in both. With V = IR, the bigger resistance has the bigger p.d.
    • If you chose B: With the same current, V = IR gives more p.d. across the larger resistance, not the same.
    • If you chose D: The sharing depends on the resistances alone. The battery p.d. only sets how big the shares are.

  7. From an earlier chapter

    What happens to an LDR's resistance when bright light falls on it?

    Tap your answer.

    1. Yes, that is right. The resistance of an LDR is high in the dark and falls as the light gets brighter.

    2. Not this one. It is the other way round. An LDR has less resistance in bright light. Go to S4.12.4. Try another.

    3. Not this one. Its resistance depends on the light, which is why it can sense it. Go to S4.12.4. Try another.

    See the answer

    It falls

    The resistance of an LDR is high in the dark and falls as the light gets brighter.

    • If you chose B: It is the other way round. An LDR has less resistance in bright light. Go to S4.12.4.
    • If you chose C: Its resistance depends on the light, which is why it can sense it. Go to S4.12.4.

  8. Three resistors of 3 Ω, 14 Ω and 22 Ω are joined in series. Find the total resistance.

    See the answer

    39 Ω

    R = 3 Ω + 14 Ω + 22 Ω = 39 Ω.

  9. A 10 V battery is joined to a 15 Ω resistor and a 35 Ω resistor in series. Find the p.d. across the 35 Ω resistor.

    See the answer

    7 V

    R = 15 Ω + 35 Ω = 50 Ω. I = 10 V ÷ 50 Ω = 0.2 A. V = IR = 0.2 A × 35 Ω = 7 V.

Keep it

Copy this into your exercise book.

Series: one path. Same current everywhere. R = R₁ + R₂. The p.d.s add up to the supply. The bigger resistor takes the bigger share. One break stops all.

Draw: A loop with a battery and two resistors in series. Write the same I on both, V₁ and V₂ across them, and V = V₁ + V₂.

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