8867 / 2027
Circuits overview

Lesson 5 of 6

Reduce a network and work back

  • I can calculate the combined resistance of two or more series resistors.Syllabus

    Syllabus 8867, 9(h). Calculate the combined resistance of two or more series resistors.

  • I can calculate the combined resistance of two or more parallel resistors.Syllabus

    Syllabus 8867, 9(i). Calculate the combined resistance of two or more parallel resistors.

  • I can solve one-source series/parallel circuits, including potential dividers using NTC thermistors and LDRs.Syllabus

    Syllabus 8867, 9(j). Solve one-source series/parallel circuits, including potential dividers using NTC thermistors and LDRs.

Find the equivalent resistance to obtain total current, then return to the original branches. A single equivalent resistor does not preserve every branch current or voltage.

Derive the two combination rules

In a series path, steady charge flow gives the same I through every resistor. The total p.d. is the sum of the individual drops:

V = IR1 + IR2 + …
Rseries = R1 + R2 + …

In parallel, each branch shares the same endpoints and therefore the same V. Total current is the sum of branch currents:

I = V/R1 + V/R2 + …
1/Rparallel = 1/R1 + 1/R2 + …

For exactly two parallel resistors, this rearranges to R1R2/(R1 + R2). Use the reciprocal sum for two or more: 3 Ω, 6 Ω and 6 Ω in parallel give 1/R = 1/3 + 1/6 + 1/6, hence R = 1.5 Ω.

Current conservation at a junction expresses conservation of charge; voltage drops along a path describe energy transferred per charge. Resistors transfer energy without consuming charge. A parallel equivalent must be smaller than the smallest positive branch resistance, while a series equivalent exceeds every individual positive resistance.

Worked network

Keep internal resistance in the complete path

A source has e.m.f. 9.0 V and internal resistance 1.0 Ω. The external circuit has R1 = 2.0 Ω on A/B, with R2 = 6.0 Ω and R3 = 3.0 Ω both on B/C.

Account for the whole circuit

Account for the whole circuitThe nine-volt ideal source raises potential from C to D. One-ohm internal resistance joins D to A, a two-ohm series resistor joins A to B, and the six-ohm and switched three-ohm branches join B to C. The switch is closed. Source current is 1.8 ampere and splits into 0.6 ampere through six ohms and 1.2 ampere through three ohms. A is at 7.2 volts and B at 3.6 volts relative to C.+-DABCr: 1 ΩR1: 2 Ω6 Ω3 ΩE = 9 VSource0.6 A1.2 A1.8 AA: 7.2 V; B: 3.6 V; C: 0 V

The branch currents add to 1.8 A. Both parallel branches have 3.6 V across the same B/C endpoints.

Open only the three-ohm branch

Open only the three-ohm branchThe nine-volt ideal source raises potential from C to D. One-ohm internal resistance joins D to A, a two-ohm series resistor joins A to B, and the six-ohm and switched three-ohm branches join B to C. Only the three-ohm branch switch is open. Source current and six-ohm branch current are both one ampere; A is at eight volts and B at six volts relative to C.+-DABCr: 1 ΩR1: 2 Ω6 Ω3 ΩE = 9 VSource1.0 AOpen: I = 01.0 AA: 8.0 V; B: 6.0 V; C: 0 V

The surviving branch receives 6.0 V and carries 1.0 A. Opening its parallel neighbour changes the whole circuit, including the internal drop.

The complete and open-branch views keep the same source and named nodes. Only the 3 ohm branch is broken in the second state; the remaining 6 ohm path still connects B to C.
RBC = (6.0 × 3.0)/(6.0 + 3.0) = 2.0 Ω
Rexternal = 2.0 + 2.0 = 4.0 Ω
I = 9.0/(4.0 + 1.0) = 1.8 A

Now work back through the original network:

VAC = 9.0 - (1.8)(1.0) = 7.2 V
VAB = (1.8)(2.0) = 3.6 V
VBC = 7.2 - 3.6 = 3.6 V
I6 = 3.6/6.0 = 0.60 A
I3 = 3.6/3.0 = 1.20 A

The branch currents add to 1.8 A. Alternatively, the simultaneous relationships (6.0 Ω)I6 = (3.0 Ω)I3 and I6 + I3 = 1.8 A give the same two values.

With C chosen as 0 V, the model potentials are D = 9.0 V, A = 7.2 V and B = 3.6 V. The ideal source raises potential from C to D; internal resistance lowers it from D to A. This is why applying 9.0 V directly across the external network would be wrong.

Pexternal = (7.2)(1.8) = 12.96 W
Pinternal = (1.8)2(1.0) = 3.24 W
Psource = (9.0)(1.8) = 16.20 W

The two destinations add to the source power, providing an independent check on the calculation.

Try it step by step

Work it out in steps: Resistance of the parallel pair in ohm, then Total resistance of the complete path including r in ohm, then Current from the cell in A.

  1. Problem 1. A cell of e.m.f. 12 V and internal resistance 1.0 ohm is connected to a 3.0 ohm resistor in series with a parallel pair of 6.0 ohm and 3.0 ohm. Find the current from the cell.
  2. Problem 2. A cell of e.m.f. 6.0 V and internal resistance 0.50 ohm drives a 2.5 ohm resistor in series with two 4.0 ohm resistors in parallel. Find the current from the cell.
  3. Problem 3. A cell of e.m.f. 10 V and internal resistance 1.0 ohm drives a 3.0 ohm resistor in series with a parallel pair of 12 ohm and 4.0 ohm. Find the current from the cell.
Show the answer

Problem 1.

Resistance of the parallel pair in ohm: Pair: (6.0 x 3.0) / (6.0 + 3.0) = 2.0 ohm

Total resistance of the complete path including r in ohm: Total = 3.0 + 2.0 + 1.0 = 6.0 ohm

Current from the cell in A: I = 12 / 6.0 = 2.0 A

Problem 2.

Resistance of the parallel pair in ohm: Pair: two equal 4.0 ohm resistors give 2.0 ohm

Total resistance of the complete path including r in ohm: Total = 2.5 + 2.0 + 0.50 = 5.0 ohm

Current from the cell in A: I = 6.0 / 5.0 = 1.2 A

Problem 3.

Resistance of the parallel pair in ohm: Pair: (12 x 4.0) / (12 + 4.0) = 3.0 ohm

Total resistance of the complete path including r in ohm: Total = 3.0 + 3.0 + 1.0 = 7.0 ohm

Current from the cell in A: I = 10 / 7.0 = 1.43 A (3 sf)

Opening a branch can increase another branch's current

Open only the 3.0 Ω branch. The external path is now 2.0 + 6.0 = 8.0 Ω, giving:

I = 9.0/(8.0 + 1.0) = 1.0 A
VAC = 9.0 - 1.0 = 8.0 V
VAB = (1.0)(2.0) = 2.0 V
VBC = 8.0 - 2.0 = 6.0 V

Total current falls from 1.8 to 1.0 A, yet current in the 6.0 Ω branch rises from 0.60 to 1.0 A. Its p.d. has increased from 3.6 to 6.0 V. The original branch voltage was not fixed independently of the rest of this circuit.

For another network, first establish what remains connected and which source model is supplied. Keep total, branch and component readings separate; a statement that one current falls does not determine every other current.

Check your understanding

A 9.0 V source with 1.0 ohm internal resistance feeds a shared 2.0 ohm resistor, then parallel 6.0 and 3.0 ohm branches. Initially total current is 1.80 A and the 6.0 ohm branch carries 0.600 A. What happens when only the 3.0 ohm branch is opened?

  1. Total current falls to 1.0 A; the 6.0 ohm branch rises to 1.0 A.
  2. Total current stays 1.80 A, and the whole of it moves into the 6.0 ohm branch.
  3. Total current becomes 0.600 A and the remaining branch stays at 0.600 A.
Show the answer

Total current falls to 1.0 A; the 6.0 ohm branch rises to 1.0 A.

Only one external route remains. Its complete resistance including the source is 9.0 ohm, giving 1.0 A. The smaller common drops leave 6.0 V across the remaining branch, rather than its previous 3.60 V.