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Parallel circuit

Curricula: KS3, GCSE (Resistance), cycle 4, 2de, Sek I, AP2, IB B.5

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Aim

Check that the total current equals the sum of the currents in the branches, and find the total resistance of the branches.

The bench

A 6 V source and an ammeter for the total current. At the nodes (buses) the circuit splits into two branches: at the top an ammeter and a 100 Ω resistor, at the bottom an ammeter and a 220 Ω resistor.

Procedure

  1. Write down the readings of the three ammeters.
  2. Compare the total current with the sum of the currents in the branches.
  3. For each branch calculate the voltage \(U = I \cdot R\).
  4. Calculate the total resistance \(R = U / I\) from the voltage in step 3 and the total current. Compare it with \(R_1 \cdot R_2 / (R_1 + R_2)\).
  5. In the settings of the lower resistor change 220 Ω to 470 Ω and take the readings again.
Expected result

The total current is about 86.3 mA; in the 100 Ω branch it is about 59.3 mA, in the 220 Ω branch about 27.0 mA; 59.3 + 27.0 = 86.3 mA. The voltage across both branches is the same: about 5.93 V. The total resistance is 5.93 V / 86.3 mA ≈ 68.7 Ω; by the formula 100 · 220 / 320 = 68.75 Ω — less than the smaller of the resistors. With the 470 Ω resistor the current in its branch falls to about 12.6 mA; the current in the 100 Ω branch hardly changes.

Questions

  1. Why did the current in the 100 Ω branch hardly change when the other resistor was replaced?
  2. How will the total current change if a third branch is added?
  3. Why is the total resistance of parallel branches less than the resistance of each of them?
  4. What is the total resistance of two 100 Ω resistors connected in parallel?
Answers
  1. Each branch is connected to the same two buses, so its current depends only on the voltage between them and on its own resistance: \(I = U / R\) ≈ 5.93 V / 100 Ω ≈ 59.3 mA. With 470 Ω the total current falls by about 14 mA, the voltage lost inside the source and the ammeter falls only by about 0.01 V, and the voltage on the buses stays almost the same.
  2. The total current increases by the current of the new branch: \(I = I_1 + I_2 + I_3\). The currents in the other branches stay almost the same, because the voltage across them hardly changes.
  3. Each branch is an extra path for the current, so at the same voltage the total current is larger than the current in any one branch, and \(R = U / I\) is smaller than the resistance of any branch. In terms of conductance, \(1 / R = 1 / R_1 + 1 / R_2\) is larger than each of its terms.
  4. \(R = R_1 \cdot R_2 / (R_1 + R_2)\) = 100 · 100 / 200 = 50 Ω, half the resistance of one resistor.