Equivalent resistance between A and B
(2 + 4) Ω in parallel with 6 Ω is 3 Ω, then 5 Ω in series, then 3 Ω and 6 Ω in parallel give 2 Ω: the resistance between A and B is 3 + 5 + 2 = 10 Ω.
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Equivalent resistance10 Ωbetween A and B
Current1 Afrom the cell
Power10 WP = VI
| R | I | V | Power | |
|---|---|---|---|---|
| R₁ | 2 Ω | 0.5 A | 1 V | 0.5 W |
| R₂ | 4 Ω | 0.5 A | 2 V | 1 W |
| R₃ | 6 Ω | 0.5 A | 3 V | 1.5 W |
| R₄ | 5 Ω | 1 A | 5 V | 5 W |
| R₅ | 3 Ω | 0.67 A | 2 V | 1.33 W |
| R₆ | 6 Ω | 0.33 A | 2 V | 0.67 W |
Step by step
- R₁ and R₂ in series: 2 + 4 = 6 Ω
- (R₁ + R₂) and R₃ in parallel: R = R₁R₂ / (R₁ + R₂) = 6 × 6 / (6 + 6) = 3 Ω
- R₅ and R₆ in parallel: R = R₁R₂ / (R₁ + R₂) = 3 × 6 / (3 + 6) = 2 Ω
- The groups are in series: Req = 3 + 5 + 2 = 10 Ω
- Current from the cell: I = V / Req = 10 / 10 = 1 A
- Across the parallel group V = IR = 1 × 3 = 3 V, the same for every branch. It splits (R₁ + R₂): 3 / 6 = 0.5 A, R₃: 3 / 6 = 0.5 A.
- Across the parallel group V = IR = 1 × 2 = 2 V, the same for every branch. It splits R₅: 2 / 3 = 0.67 A, R₆: 2 / 6 = 0.33 A.
- Drops in series: R₄: 1 × 5 = 5 V.
- Power: P = EI = 10 × 1 = 10 W.
Why it works
- Mark the two terminals A and B. Every path from A to B carries part of the current.
- Resistors one after another on a single path, with no junction between them, are in series: add them.
- Resistors that join the same two junctions are in parallel: add their reciprocals.
- Replace each group by its equivalent and redraw. Repeat until one resistor joins A and B.
- If a network will not reduce this way, as with an unbalanced bridge or a cube, use symmetry or Kirchhoff's laws.
Resistance between A and B: quick answers
How do you find the equivalent resistance between two points?
Reduce series and parallel groups one at a time, working inwards from the parts farthest from A and B, until a single resistor is left between them.
What is the resistance between A and B here?
10 Ω. With 10 V across A and B the current is 1 A, and the 5 Ω resistor in the middle drops 5 V.
What if the network is a bridge?
A balanced Wheatstone bridge carries no current in its middle arm, so that arm can be removed. An unbalanced one needs Kirchhoff's laws.