Find the value of current I drawn from the cell.
Calculate the current I.
Topic: Series-parallel resistance
Answer
From the circuit, two 15 Ω resistors are in series in one branch, and that branch is in parallel with a 30 Ω resistor. The cell has e.m.f. 3.4 V and internal resistance 2 Ω.
Step 1 — the series pair: R₁ = 15 + 15 = 30 Ω
Step 2 — in parallel with the 30 Ω: R_ext = (30 × 30) ÷ (30 + 30) = 900 ÷ 60 = 15 Ω
Step 3 — add the internal resistance: total = 15 + 2 = 17 Ω
Step 4 — the current: I = ε ÷ (R_ext + r) = 3.4 ÷ 17 = 0.2 A
Current drawn from the cell, I = 0.2 A
Reduce the network first, then add the internal resistance, then apply Ohm's law — in that order, every time.
Two equal resistors in parallel give exactly half of one of them, so 30 Ω and 30 Ω give 15 Ω. Worth spotting rather than grinding through the formula.
The internal 2 Ω must go into the division. Using 3.4 ÷ 15 gives 0.227 A, the standard error whenever a cell has internal resistance.