Chapter 6 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

S4.12.6 Chapter 6

Parallel circuits

Ana joins two lamps side by side across one battery. Both glow as brightly as one lamp, and she can unscrew one while the other stays lit. How is this loop different?

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

  • Explain how current splits between parallel branches
  • Explain why adding resistors in parallel lowers the total resistance

In a parallel circuit the current has more than one path. Each path is a branch. The branches start at one junction and meet again at another, and each branch is joined straight across the supply.

Take a 6 V battery with two branches. One branch has a 6 Ω resistor and the other has a 12 Ω resistor. An ammeter is in each branch, one more is on the wire from the battery, and a voltmeter is across each resistor.

A battery labelled 6 V with an ammeter on the wire from it. The wire splits at a junction into two branches that meet again. The first branch has an ammeter that reads 1 A and a resistor labelled 6 Ω. The second branch has an ammeter that reads 0.5 A and a resistor labelled 12 Ω. A voltmeter across each resistor reads 6 V. The ammeter on the battery wire reads 1.5 A. Labels: 6 V, A, 1.5 A, A, 1 A, 6 Ω, V, 6 V, A, 0.5 A, 12 Ω, V, 6 V.6 VA1.5 AA1 AV6 V6 ΩA0.5 AV6 V12 Ω
A parallel circuit. Both voltmeters read 6 V. The branch currents are 1 A and 0.5 A, and the battery gives 1.5 A.

Rule 1: every branch has the same p.d. as the supply. Each branch is joined to the same two points. Each coulomb that takes a branch gains the battery's energy, and gives all of it away in that branch.

Rule 2: the currents in the branches add up to the current from the battery. Charge is not lost at a junction, so what goes in comes out. I = I₁ + I₂. Here 1 A + 0.5 A = 1.5 A.

Each branch is a loop of its own with the battery, so each follows V = IR. The branch with 6 Ω has I = 6 V ÷ 6 Ω = 1 A. The branch with 12 Ω has I = 6 V ÷ 12 Ω = 0.5 A. The smaller resistance takes the bigger current, in inverse proportion to the resistance.

See it three ways

As a picture

A table with the columns Quantity and In parallel. Row 1: p.d., same. Row 2: current, adds up. Row 3: resistance, less than all.QuantityIn parallelp.d.samecurrentadds upresistanceless than all

In words

Each branch has the full p.d. The branch currents add up to the current from the battery. The total resistance is less than the smallest branch.

In symbols

V = V₁ = V₂. I = I₁ + I₂. R is less than R₁ and less than R₂.

Now the surprising part. The total resistance of the two branches is smaller than either resistor. Think of paths. Each new branch gives the charge another way through. The p.d. is unchanged, so the old branches carry the same current as before, and the new branch adds its own. Watch it with 6 Ω resistors on a 6 V battery.

Draw it first

More branches, more current

  1. A battery labelled 6 V with an ammeter. Step 1: one resistor labelled 6 Ω. The ammeter reads 1 A. Step 2: a second 6 Ω resistor is joined in parallel, as a second branch. The ammeter reads 2 A. Step 3: a third 6 Ω resistor is added as a third branch. The ammeter reads 3 A. Labels: 6 V, A, 1 A, 6 Ω, 2 A, 6 Ω, 3 A, 6 Ω.6 VA1 A6 Ω

    1 One 6 Ω resistor. The battery gives I = 6 V ÷ 6 Ω = 1 A.

  2. A battery labelled 6 V with an ammeter. Step 1: one resistor labelled 6 Ω. The ammeter reads 1 A. Step 2: a second 6 Ω resistor is joined in parallel, as a second branch. The ammeter reads 2 A. Step 3: a third 6 Ω resistor is added as a third branch. The ammeter reads 3 A. Labels: 6 V, A, 1 A, 6 Ω, 2 A, 6 Ω, 3 A, 6 Ω.6 VA2 A6 Ω6 Ω

    2 A second branch takes another 1 A. The battery gives 2 A, so the total resistance is 6 V ÷ 2 A = 3 Ω.

  3. A battery labelled 6 V with an ammeter. Step 1: one resistor labelled 6 Ω. The ammeter reads 1 A. Step 2: a second 6 Ω resistor is joined in parallel, as a second branch. The ammeter reads 2 A. Step 3: a third 6 Ω resistor is added as a third branch. The ammeter reads 3 A. Labels: 6 V, A, 1 A, 6 Ω, 2 A, 6 Ω, 3 A, 6 Ω.6 VA3 A6 Ω6 Ω6 Ω

    3 A third branch takes another 1 A. The battery gives 3 A, so the total resistance is 6 V ÷ 3 A = 2 Ω. Each new branch lowers the total resistance.

Back to the first loop. The battery gives 1.5 A at 6 V, so the total resistance is 6 V ÷ 1.5 A = 4 Ω. That is less than 6 Ω, the smallest branch. The 12 Ω branch is an extra path, and it can only add current, never take any away. So the total is always less than the smallest branch.

There is a rule for the total resistance R of resistors in parallel. It adds the reciprocals: 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃. For just two resistors it simplifies to the product over the sum: R = (R₁ × R₂) ÷ (R₁ + R₂).

Worked example

Two resistors of 6 Ω and 12 Ω are joined in parallel. What is the total resistance?

  1. Write the rule for two resistors: R = (R₁ × R₂) ÷ (R₁ + R₂).
  2. Multiply: 6 × 12 = 72. Add: 6 + 12 = 18.
  3. Divide: R = 72 ÷ 18 = 4 Ω.

Answer: 4 Ω

Worked example

Three resistors of 10 Ω, 20 Ω and 20 Ω are joined in parallel. What is the total resistance?

  1. Write the rule: 1 ÷ R = 1 ÷ R₁ + 1 ÷ R₂ + 1 ÷ R₃.
  2. Put in the numbers: 1 ÷ R = 1 ÷ 10 + 1 ÷ 20 + 1 ÷ 20 = 0.1 + 0.05 + 0.05 = 0.2.
  3. Turn it over: R = 1 ÷ 0.2 = 5 Ω.

Answer: 5 Ω

A battery labelled 6 V with two branches joined across it. One branch has a resistor labelled 20 Ω and the other a resistor labelled 30 Ω. No meters are drawn. Labels: 6 V, 20 Ω, 30 Ω.6 V20 Ω30 Ω
The loop of the next worked example, a 6 V battery with a 20 Ω branch and a 30 Ω branch.

Worked example

A 6 V battery is joined across a 20 Ω resistor and a 30 Ω resistor in parallel. Find the current in each branch and the current from the battery.

  1. Each branch has the full 6 V across it.
  2. The 20 Ω branch: I = V ÷ R = 6 V ÷ 20 Ω = 0.3 A.
  3. The 30 Ω branch: I = V ÷ R = 6 V ÷ 30 Ω = 0.2 A.
  4. The current from the battery is the sum: 0.3 A + 0.2 A = 0.5 A.
  5. Check with the total resistance: R = (20 × 30) ÷ (20 + 30) = 12 Ω, and 6 V ÷ 12 Ω = 0.5 A.

Answer: 0.3 A, 0.2 A and 0.5 A

Homes are wired in parallel. Each lamp and each socket is on its own branch, with the full p.d. of the supply. Each works at full strength, and a switch in one branch turns off only that branch. Look at two lamps with a switch in one branch.

A battery labelled 6 V with two branches joined across it. The first branch has a lamp labelled lamp 1, which is lit. The second branch has an open switch and a lamp labelled lamp 2, which is dark. Labels: 6 V, lamp 1, lamp 2.6 Vlamp 1lamp 2
The switch in the second branch is open, so lamp 2 is off. Lamp 1 is on its own branch, and it stays lit at full brightness.

Try it

Two bulbs side by side

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 cells 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. Look at how bright it is.
  2. Now join the second bulb across the same two ends, side by side with the first. Each bulb needs its own wires to the cells. Look at both bulbs.
  3. Lift one bulb out of its place. Look at the other bulb.

What you will see: The two bulbs side by side are as bright as one on its own. Each has its own path, and each has the full p.d. of the cells. Take one away and the other stays lit, because its branch is still complete. The cells give twice the current, so they run down faster.

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 in parallel, both bright. Row 3: one lifted, one bright.LoopBulbsone bulbbrighttwo in parallelboth brightone liftedone bright

A wrong idea many people have

Resistors in parallel add up: R = R₁ + R₂.

Why it feels right: Resistors in series do add, and "total resistance" sounds like a sum.

The test that breaks it: Add 6 Ω and 12 Ω and you get 18 Ω, which is more than either. But the second branch is an extra path, and the loop gave 1.5 A from 6 V, which is 4 Ω.

What is true: In parallel the total is smaller than the smallest branch. Use the product over the sum for two resistors, and then check that your answer is smaller than both.

A wrong idea many people have

In a parallel circuit the p.d. is shared between the branches, as it is in series.

Why it feels right: We say the p.d. is shared in a series circuit, and it is natural to carry the idea over.

The test that breaks it: Look at the two voltmeters in the main picture. Each reads the full 6 V of the battery. No branch has only a part of it.

What is true: Each branch is joined across the supply, so each has the full p.d. In parallel it is the current that is shared out.

A wrong idea many people have

The current splits equally between the branches.

Why it feels right: Two paths look like two halves.

The test that breaks it: In the main picture the branches carry 1 A and 0.5 A. The branch with the smaller resistance took twice as much.

What is true: The current splits in inverse proportion to the resistances. With the same p.d., I = V ÷ R, so the smaller resistance takes the bigger current. The shares are equal only when the resistances are equal.

Key words

parallel circuit
a circuit with more than one path, where each branch is joined straight across the supply.
branch
one of the paths in a parallel circuit.

Back to your guess

The answer is C: Each lamp has its own path, and each path has the full p.d. of the battery.

In a parallel circuit each lamp is joined straight across the battery. Each has the full p.d., so each glows as brightly as it would alone. A break in one branch leaves the other branch complete.

Check yourself

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

  1. Resistors of 3 Ω and 6 Ω are joined in parallel. What is the total resistance?

    Tap your answer.

    1. Not this one. You added them. That is the rule for series. In parallel the total is smaller than the smallest branch. Try another.

    2. Not this one. You took the average. The total in parallel is less than the smallest branch, which is 3 Ω. Try another.

    3. Not this one. You multiplied but forgot to divide by the sum. The rule is the product over the sum. Try another.

    4. Yes, that is right. R = (3 × 6) ÷ (3 + 6) = 18 ÷ 9 = 2 Ω. It is less than 3 Ω, the smallest branch, as it must be.

    See the answer

    2 Ω

    R = (3 × 6) ÷ (3 + 6) = 18 ÷ 9 = 2 Ω. It is less than 3 Ω, the smallest branch, as it must be.

    • If you chose A: You added them. That is the rule for series. In parallel the total is smaller than the smallest branch.
    • If you chose B: You took the average. The total in parallel is less than the smallest branch, which is 3 Ω.
    • If you chose C: You multiplied but forgot to divide by the sum. The rule is the product over the sum.

  2. The picture shows a 12 V battery with an 8 Ω branch and a 24 Ω branch. The ammeters in the branches read 1.5 A and 0.5 A. What does the ammeter A on the wire from the battery read?

    A battery labelled 12 V with an ammeter labelled A on the wire from it. The wire splits into two branches. The first branch has an ammeter that reads 1.5 A and a resistor labelled 8 Ω. The second branch has an ammeter that reads 0.5 A and a resistor labelled 24 Ω. Labels: 12 V, A, A, 1.5 A, 8 Ω, A, 0.5 A, 24 Ω.12 VAAA1.5 A8 ΩA0.5 A24 Ω
    Tap your answer.

    1. Yes, that is right. The branch currents add up to the current from the battery: 1.5 A + 0.5 A = 2 A. Check: 12 V ÷ 2 A = 6 Ω, which is the total resistance of 8 Ω and 24 Ω in parallel.

    2. Not this one. That is the current in one branch only. The battery supplies both branches. Go to S3.10.3. Try another.

    3. Not this one. You took the average of the two. The currents add, because charge is not lost at a junction. Try another.

    4. Not this one. That is the current in the other branch. The battery supplies both branches. Go to S3.10.3. Try another.

    See the answer

    2 A

    The branch currents add up to the current from the battery: 1.5 A + 0.5 A = 2 A. Check: 12 V ÷ 2 A = 6 Ω, which is the total resistance of 8 Ω and 24 Ω in parallel.

    • If you chose B: That is the current in one branch only. The battery supplies both branches. Go to S3.10.3.
    • If you chose C: You took the average of the two. The currents add, because charge is not lost at a junction.
    • If you chose D: That is the current in the other branch. The battery supplies both branches. Go to S3.10.3.

  3. A 5 Ω resistor and a 20 Ω resistor are joined in parallel across a battery. Which one carries the bigger current?

    Tap your answer.

    1. Not this one. More resistance means less current for the same p.d. I = V ÷ R. Try another.

    2. Not this one. The p.d. is the same, so the current depends on the resistance. I = V ÷ R gives different currents. Try another.

    3. Yes, that is right. Each branch has the full p.d. I = V ÷ R, so the smaller resistance has the bigger current. Here it is four times as big.

    4. Not this one. The comparison does not need the p.d. The same p.d. is across both, so the smaller resistance has the bigger current. Try another.

    See the answer

    The 5 Ω resistor: the p.d. is the same, and its resistance is smaller

    Each branch has the full p.d. I = V ÷ R, so the smaller resistance has the bigger current. Here it is four times as big.

    • If you chose A: More resistance means less current for the same p.d. I = V ÷ R.
    • If you chose B: The p.d. is the same, so the current depends on the resistance. I = V ÷ R gives different currents.
    • If you chose D: The comparison does not need the p.d. The same p.d. is across both, so the smaller resistance has the bigger current.

  4. Two lamps are joined in parallel across a 3 V battery. What is the p.d. across each lamp?

    Tap your answer.

    1. Not this one. That is how a series circuit shares the p.d. In parallel each lamp is joined straight across the battery. Try another.

    2. Not this one. The p.d. of the battery is 3 V, and no lamp can have more than the battery gives. Try another.

    3. Yes, that is right. Each branch is joined straight across the battery, so each has the full 3 V.

    4. Not this one. Both lamps are joined across the same two points, so both have the same p.d. Try another.

    See the answer

    3 V across each lamp

    Each branch is joined straight across the battery, so each has the full 3 V.

    • If you chose A: That is how a series circuit shares the p.d. In parallel each lamp is joined straight across the battery.
    • If you chose B: The p.d. of the battery is 3 V, and no lamp can have more than the battery gives.
    • If you chose D: Both lamps are joined across the same two points, so both have the same p.d.

  5. Why is the total resistance of two resistors in parallel less than either one alone?

    Tap your answer.

    1. Not this one. Resistors do not cancel. Each one resists, but the charge has two paths to choose from. Try another.

    2. Yes, that is right. The same p.d. is across each branch, so the old branch carries the same as before and the new branch adds more. More current for the same p.d. means a smaller total resistance, R = V ÷ I.

    3. Not this one. The battery p.d. does not change. The extra current comes from the extra path. Try another.

    4. Not this one. Current goes through both branches. The smaller resistance takes more of it. Try another.

    See the answer

    The charge has two paths, so more current flows than through one alone

    The same p.d. is across each branch, so the old branch carries the same as before and the new branch adds more. More current for the same p.d. means a smaller total resistance, R = V ÷ I.

    • If you chose A: Resistors do not cancel. Each one resists, but the charge has two paths to choose from.
    • If you chose C: The battery p.d. does not change. The extra current comes from the extra path.
    • If you chose D: Current goes through both branches. The smaller resistance takes more of it.

  6. A house is wired with its lamps in parallel. What is the main reason?

    Tap your answer.

    1. Yes, that is right. Each lamp is on a branch across the supply, so it has the full p.d. and works at full strength. A switch or a break in one branch leaves the others complete.

    2. Not this one. In parallel the p.d. is the same for each branch, and the currents can differ. Try another.

    3. Not this one. Parallel branches make the total resistance smaller, and more branches draw more current. Try another.

    4. Not this one. It is series lamps that share the p.d. In parallel each lamp has the full p.d. Go to S3.10.3. Try another.

    See the answer

    Each lamp has the full p.d., so one off or broken leaves the rest lit

    Each lamp is on a branch across the supply, so it has the full p.d. and works at full strength. A switch or a break in one branch leaves the others complete.

    • If you chose B: In parallel the p.d. is the same for each branch, and the currents can differ.
    • If you chose C: Parallel branches make the total resistance smaller, and more branches draw more current.
    • If you chose D: It is series lamps that share the p.d. In parallel each lamp has the full p.d. Go to S3.10.3.

  7. From an earlier chapter

    A 12 V battery is joined to an 18 Ω resistor and a 30 Ω resistor in series. What current flows?

    Tap your answer.

    1. Not this one. You used only the 18 Ω resistor. In series, add the two resistances first. Go to S4.12.5. Try another.

    2. Not this one. You used only the 30 Ω resistor. Both are in the loop, so use the total. Go to S4.12.5. Try another.

    3. Not this one. You divided the total resistance by the p.d. Use I = V ÷ R. Go to S4.12.5. Try another.

    4. Yes, that is right. In series R = 18 Ω + 30 Ω = 48 Ω. I = V ÷ R = 12 V ÷ 48 Ω = 0.25 A.

    See the answer

    0.25 A

    In series R = 18 Ω + 30 Ω = 48 Ω. I = V ÷ R = 12 V ÷ 48 Ω = 0.25 A.

    • If you chose A: You used only the 18 Ω resistor. In series, add the two resistances first. Go to S4.12.5.
    • If you chose B: You used only the 30 Ω resistor. Both are in the loop, so use the total. Go to S4.12.5.
    • If you chose C: You divided the total resistance by the p.d. Use I = V ÷ R. Go to S4.12.5.

  8. Two resistors of 20 Ω and 60 Ω are joined in parallel. Find the total resistance.

    See the answer

    15 Ω

    R = (20 × 60) ÷ (20 + 60) = 1 200 ÷ 80 = 15 Ω.

  9. A 6 V battery is joined to a 12 Ω resistor and an 8 Ω resistor in parallel. Find the current from the battery.

    See the answer

    1.25 A

    The 12 Ω branch: 6 V ÷ 12 Ω = 0.5 A. The 8 Ω branch: 6 V ÷ 8 Ω = 0.75 A. The battery gives 0.5 A + 0.75 A = 1.25 A.

Keep it

Copy this into your exercise book.

Parallel: branches across the supply. Same p.d. in each. Branch currents add. Smaller R takes more current. Total R is less than the smallest. Two resistors: R = R₁R₂ ÷ (R₁ + R₂).

Draw: A battery with two branches, each with a resistor. Write the same V on both, I₁ and I₂ on the branches and I = I₁ + I₂ at the junction.

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