Worked examples (new values)

Example 1 — series. Resistors of 10 Ω, 20 Ω and 30 Ω are in series with a 30 V battery of negligible internal resistance. R = 10 + 20 + 30 = 60 Ω. I = 30/60 = 0.5 A in every resistor. V₁ = 0.5 × 10 = 5 V, V₂ = 0.5 × 20 = 10 V, V₃ = 0.5 × 30 = 15 V; the sum is 30 V, equal to the battery emf.

Example 2 — parallel. The same three resistors are connected in parallel across the same battery. Each has 30 V across it: I₁ = 3 A, I₂ = 1.5 A, I₃ = 1 A, total 5.5 A. 1/R = 1/10 + 1/20 + 1/30 = (6 + 3 + 2)/60 = 11/60, so R = 60/11 = 5.45 Ω. Check: 30 / 5.45 = 5.5 A.

Example 3 — mixed. A 4 Ω and a 12 Ω resistor are in parallel; this pair is in series with a 5 Ω resistor and a 16 V battery (r = 0). Parallel pair: 4×12 / 16 = 3 Ω. Total R = 3 + 5 = 8 Ω. I = 16/8 = 2 A. Voltage across the pair = 2 × 3 = 6 V. So I(4 Ω) = 6/4 = 1.5 A and I(12 Ω) = 6/12 = 0.5 A (sum = 2 A).

Example 3a — a second series circuit. Three resistors of 8 Ω, 15 Ω and 27 Ω are joined in series across a 60 V supply (negligible internal resistance). (a) R = 8 + 15 + 27 = 50 Ω. (b) I = 60 / 50 = 1.2 A, the same in all three. (c) V₁ = 1.2 × 8 = 9.6 V, V₂ = 1.2 × 15 = 18 V, V₃ = 1.2 × 27 = 32.4 V. Sum = 60 V ✔. Notice that the 27 Ω resistor, being the largest, takes the largest share (54%) of the voltage.

Example 3b — a second parallel circuit. Resistors of 12 Ω, 20 Ω and 30 Ω are connected in parallel across a 24 V battery with negligible internal resistance. (a) Each branch has 24 V: I₁ = 24/12 = 2 A, I₂ = 24/20 = 1.2 A, I₃ = 24/30 = 0.8 A. (b) The main current is I = 2 + 1.2 + 0.8 = 4 A. (c) 1/R = 1/12 + 1/20 + 1/30 = (5 + 3 + 2)/60 = 10/60, so R = 6 Ω; check: 24/6 = 4 A ✔. (d) R = 6 Ω is smaller than the smallest resistor (12 Ω), as it must be.

Example 3c — a mixed circuit with the parallel group at the end. A 20 Ω and a 5 Ω resistor are in parallel; this pair is in series with a 6 Ω resistor, and the combination is connected to a 15 V battery with negligible internal resistance. Step 1: parallel pair = 20 × 5 / 25 = 4 Ω. Step 2: R = 4 + 6 = 10 Ω. Step 3: I = 15 / 10 = 1.5 A (this is also the current in the 6 Ω resistor). Step 4: voltage across the pair = 1.5 × 4 = 6 V. Step 5: I(20 Ω) = 6 / 20 = 0.3 A; I(5 Ω) = 6 / 5 = 1.2 A; their sum is 1.5 A ✔.

Example 3d — a series group inside a parallel group. A 6 Ω and a 12 Ω resistor in series form one branch; this branch is in parallel with a 9 Ω resistor; the pair of branches is in series with a 4 Ω resistor, across a 20 V battery of negligible internal resistance. Step 1: the series branch = 6 + 12 = 18 Ω. Step 2: parallel with 9 Ω = 18 × 9 / 27 = 6 Ω. Step 3: R = 6 + 4 = 10 Ω, so I = 20 / 10 = 2 A. Step 4: voltage across the parallel part = 2 × 6 = 12 V; the 4 Ω resistor takes 8 V (12 + 8 = 20 ✔). Step 5: I in the 18 Ω branch = 12 / 18 = 0.667 A; I in the 9 Ω resistor = 12 / 9 = 1.333 A; sum = 2 A ✔. Step 6: the voltage across the 6 Ω resistor is 0.667 × 6 = 4 V and across the 12 Ω resistor is 8 V.

Example 3e — equal resistors. Five identical 10 Ω resistors are connected (i) in series and (ii) in parallel. (i) R = 5 × 10 = 50 Ω. (ii) R = 10 / 5 = 2 Ω. The ratio of the two combinations is 25 : 1.