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Series circuits

Electricity and magnetism · level 5, lesson 3 of 6

A question to start

Two bulbs in one loop are dimmer than one on its own. Can you work out exactly how much current flows?

The idea

Parts in series are in one single loop, one after another.

  • The current is the same through every part.
  • The resistances add: \(R = R_1 + R_2 + \dots\)
  • The battery's voltage is shared between the parts, in proportion to their resistances: \(V_1 = IR_1\), and so on.

So adding a part in series always increases the total resistance and lowers the current.

2 Ω4 Ω4 V8 V12 VI = 2 A

Worked examples

1. The diagram. \(R = 2 + 4 = 6\ \Omega\). \(I = 12/6 = 2\) A. \(V_1 = 2 \times 2 = 4\) V and \(V_2 = 2 \times 4 = 8\) V; \(4 + 8 = 12\) V.

2. Two bulbs. Each bulb is about 12 \(\Omega\), on 3 V. One bulb: \(I = 0.25\) A. Two in series: \(I = 3/24 = 0.125\) A, half as much, so both are dimmer.

3. Limiting an LED. An LED needs 2 V and 0.02 A, from a 9 V battery. The resistor must take \(9 - 2 = 7\) V: \(R = 7/0.02 = 350\ \Omega\).

Try it

  1. Find the total resistance of 5 \(\Omega\), 10 \(\Omega\) and 15 \(\Omega\) in series.
  2. Those three are put across 60 V. Find the current and the voltage across each.
  3. Why does adding a bulb in series make the others dimmer?
  4. An LED needs 3 V and 0.02 A from a 5 V supply. What resistor is needed?
Answers
  1. \(30\ \Omega\).
  2. \(I = 60/30 = 2\) A; 10 V, 20 V and 30 V.
  3. The total resistance goes up, so the current through all of them goes down.
  4. \((5 - 3)/0.02 = 100\ \Omega\).

Common mistake

"The current is used up along a series loop." It is the same all the way round. The voltage is what is shared out.

Where it leads