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Series circuit

Curricula: KS3, GCSE (Resistance), cycle 4, Sek I, AP2, IB B.5

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Aim

Check that in a series circuit the current is the same everywhere, and that the voltages across the parts and the resistances add up.

The bench

A 9 V source, an ammeter and 100 Ω, 220 Ω and 470 Ω resistors connected in series. Above each resistor is its own voltmeter.

Procedure

  1. Write down the readings of the ammeter and the three voltmeters.
  2. Add up the voltages across the resistors and compare the sum with the source voltage.
  3. For each resistor calculate the current \(I = U / R\) and compare it with the ammeter reading.
  4. Calculate the total resistance \(R = (U_1 + U_2 + U_3) / I\) and compare it with the sum \(R_1 + R_2 + R_3\).
  5. Reduce the source voltage to 4.5 V and repeat.
Expected result

The current is about 11.4 mA. The voltages are 1.14 V, 2.50 V and 5.35 V; they add up to about 8.99 V — almost the whole source voltage. \(U / R\) for all three resistors gives the same 11.4 mA. The total resistance is 8.99 V / 11.4 mA ≈ 790 Ω — the sum 100 + 220 + 470 Ω. The voltages divide in proportion to the resistances. At 4.5 V the voltages and the current are half as large, while their ratios and the total resistance are the same.

Questions

  1. Which resistor has the greatest voltage across it, and why?
  2. What is the total resistance of two 100 Ω resistors connected in series?
Answers
  1. The 470 Ω resistor, about 5.35 V. The same current of about 11.4 mA flows through all three resistors, so by \(U = I \cdot R\) the voltage is largest across the largest resistance.
  2. In series the resistances add up: \(R = R_1 + R_2\) = 100 + 100 = 200 Ω.