Series & Parallel
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.
Series: one path, so the current is the same and the voltages add up. Parallel: separate branches between the same two points, so the voltage is the same and the currents add up. Adding a parallel branch always lowers the total resistance.
1. Combining resistors
2. Solving a circuit
1. Collapse the circuit to one Req, working from the inside out.
2. Find the battery current: I = ε/Req.
3. Work back out: series parts share I, parallel parts share ΔV.
In the figure: 6 Ω ∥ 3 Ω = 2 Ω, plus 4 Ω gives 6 Ω. I = 18/6 = 3 A. The 4 Ω takes 12 V, leaving 6 V for the pair: 6/6 = 1 A and 6/3 = 2 A. Check: 1 + 2 = 3 A ✓.
3. Example, then try it
Two 10 Ω resistors are connected to a 20 V battery, first in series, then in parallel. Find the battery current each time.
Series: Req = 20 Ω, I = 1.0 A.
Parallel: Req = 5.0 Ω, I = 4.0 A. Parallel draws more current.
Three identical 4.0 Ω bulbs run on a 12 V battery: bulb A in series with B and C, which are in parallel with each other. Find the power of each bulb. Then bulb C is unscrewed. What happens to A and B?
Before: B ∥ C = 2.0 Ω, so Req = 6.0 Ω and I = 2.0 A through A, splitting into 1.0 A each through B and C.
Power: A: (2.0)²(4.0) = 16 W; B and C: (1.0)²(4.0) = 4.0 W each. A is brightest.
Unscrew C: A and B are now in series: Req = 8.0 Ω, I = 1.5 A.
After: A and B each get (1.5)²(4.0) = 9.0 W. A gets dimmer (16 → 9 W) and B gets brighter (4 → 9 W).
Answer: A 16 W, B and C 4 W each; then A dims to 9 W and B brightens to 9 W.
4. Common mistakes
A 3 Ω resistor and a 6 Ω resistor are connected in parallel. What is their equivalent resistance?
(3)(6) ÷ (3 + 6) = 18 ÷ 9 = 2 Ω, smaller than either resistor.
Ready to practice?
6 questions on Series & Parallel, each with a hint and an explanation for every wrong answer.