Parallel circuits

In a parallel circuit there are branches. Each branch gets the full supply p.d., the branch currents add up to the total, and adding a branch lowers the total resistance. Homes are wired in parallel so each appliance works on its own.

Key facts

  • Parallel: each branch has the full supply p.d.; currents add: I = I₁ + I₂ + …
  • Parallel resistors: 1/R = 1/(R₁) + 1/(R₂) + … (for two: product ÷ sum). The total is less than the smallest.
  • If one bulb breaks in parallel, only its branch stops: the others stay on.
  • Adding another resistor in parallel lowers the total resistance, because it adds another path for the current.

Worked examples

Example 1

The ammeters in the branches read 0.4 A and 0.7 A. What does the ammeter by the battery read?

  1. The current from the battery splits at the junction: part goes through each branch.
  2. The branch currents join again, so they add up to the total. Total = 0.4 + 0.7 = 1.1 A

Answer: 1.1 A

Example 2

In a parallel circuit the ammeter next to the cell reads 0.9 A. The ammeter in one branch reads 0.5 A. What does the ammeter in the other branch read?

  1. In a parallel circuit the current splits between the branches, and joins up again afterwards.
  2. So the branch currents add up to the current from the cell. 0.9 − 0.5 = 0.4 A

Answer: 0.4 A

Example 3

One bulb breaks in a parallel circuit of three bulbs. What happens to the other two?

  1. In a parallel circuit, each bulb is on its own branch.
  2. The broken bulb only breaks its own branch, so current still flows through the others.

Answer: They stay on.

Example 4

Why are the lights and sockets in a house wired in parallel?

  1. In parallel, each branch gets the full supply p.d.
  2. Each branch can have its own switch, and if one appliance fails the others still work.

Answer: Each one gets the full mains p.d. and can be switched on and off on its own.

Example 5

Two lamps are in parallel with a 6 V battery. What is the p.d. across each lamp?

  1. Each branch is connected straight across the battery.
  2. So each lamp has the full 6 V across it.

Answer: 6 V

Example 6

Two identical lamps are in parallel. The ammeter next to the battery reads 1.5 A. What current flows through each lamp?

  1. The branch currents add up to the battery current.
  2. The lamps are identical, so the current splits equally. 1.5 ÷ 2 = 0.75 A

Answer: 0.75 A

Example 7

A 12 V battery is connected to 4 Ω and 6 Ω resistors in parallel. Find the current through each resistor and the battery current.

  1. Each resistor has the full 12 V across it.
  2. Use I = V ÷ R for each branch. 12 ÷ 4 = 3 A, 12 ÷ 6 = 2 A
  3. The branch currents add up. 3 + 2 = 5 A

Answer: 3 A and 2 A; 5 A from the battery

Example 8

Adding another resistor in parallel does what to the total resistance?

  1. A resistor in parallel adds another path for the current.
  2. More current flows from the battery for the same p.d., so the total resistance is smaller.

Answer: It goes down.

Example 9

Resistors of 10 Ω and 40 Ω are in parallel across 8 V. Which one carries more current, and how much more?

  1. Both resistors have the full 8 V across them.
  2. Find each current. 8 ÷ 10 = 0.8 A, 8 ÷ 40 = 0.2 A
  3. The smaller resistor carries more current. 0.8 − 0.2 = 0.6 A

Answer: the 10 Ω resistor; 0.6 A more

Example 10

Find the combined resistance of 6 Ω and 3 Ω in parallel.

  1. For two resistors in parallel, use product ÷ sum.
  2. Put in the values. (6 × 3) ÷ (6 + 3) = 18 ÷ 9 = 2 Ω
  3. Check: 2 Ω is less than the smallest resistor, 3 Ω.

Answer: 2 Ω

Example 11

Resistors of 10 Ω, 20 Ω and 20 Ω are in parallel. Find the total resistance.

  1. For resistors in parallel, add the reciprocals. 1/R = 1/10 + 1/20 + 1/20 = 4/20
  2. Turn it upside down. R = 20 ÷ 4 = 5 Ω

Answer: 5 Ω

Example 12

You need a combined resistance of 4 Ω but only have a 12 Ω resistor. What resistor must you connect in parallel with it?

  1. In parallel, the reciprocals add. 1/4 = 1/12 + 1/R
  2. Rearrange. 1/R = 1/4 − 1/12 = 2/12 = 1/6
  3. Turn it upside down. R = 6 Ω

Answer: 6 Ω