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Kirchhoff's rules

Curricula: AP2, APC, A-level, 2de, Sek II

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Aim

Verify Kirchhoff's first and second rules in a circuit with two sources.

The bench

At the top a 6 V source, an ammeter and a 100 Ω resistor; at the bottom a 3 V source, an ammeter, a 220 Ω resistor and a third branch — an ammeter and a 330 Ω resistor. The branches meet at junction A and return to the negative terminals of the sources through junction B (buses).

Procedure

  1. Note the readings of the three ammeters. The ammeter of the lower source is connected so that it shows the current flowing into the 3 V source from the side of junction A.
  2. Verify the first rule for junction A: the current coming from the 6 V source splits between the two other branches, \(I_1 = I_2 + I_3\).
  3. Calculate the voltages across the resistors, \(U = I \cdot R\), and verify the second rule for each loop.
  4. Set 6 V on the lower source and repeat.
Expected result

\(I_1\) ≈ 18.0 mA, \(I_2\) ≈ 5.4 mA, \(I_3\) ≈ 12.7 mA; 5.4 + 12.7 ≈ 18.0 mA. The current \(I_2\) flows into the 3 V source against its EMF: the stronger 6 V source charges it. The loop with the 6 V source: 1.80 V + 4.18 V ≈ 6 V. The loop with the 3 V source and the 330 Ω resistor: 3 V + 1.19 V across 220 Ω ≈ 4.18 V across 330 Ω. With 6 V on the lower source the current in its branch reverses direction, and the ammeter shows about −4.7 mA; \(I_1\) ≈ 10.3 mA, \(I_3\) ≈ 15.0 mA, and again \(I_1 = I_2 + I_3\).

Questions

  1. Why does the current through the 3 V source flow against its EMF?
  2. At what voltage of the lower source will the current in its branch become zero?
Answers
  1. The 6 V source keeps junction A about 4.18 V above junction B: that is the voltage across the 330 Ω resistor. It is higher than the 3 V EMF of the lower source, so the difference, about 1.2 V across the 220 Ω resistor, drives about 5.4 mA from A into the positive terminal of the 3 V source.
  2. The current is zero when there is no voltage across the 220 Ω resistor, that is, when the lower source equals the voltage at junction A. With \(I_2 = 0\) the 6 V source drives \(I_1 = I_3\) through 100 Ω and 330 Ω in series, and A is at 6 V · 330 / 430 ≈ 4.6 V, so the lower source must be set to about 4.6 V.