QuestionA 9.0 V battery of negligible internal resistance is connected to a 2.0 Ω resistor R1 in series with a parallel pair, R2 = 6.0 Ω and R3 = 12 Ω. Find the current through, and the p.d. across, every resistor.
Rparallel = (6.0 × 12)/(6.0 + 12) = 4.0 Ω Rtotal = 2.0 + 4.0 = 6.0 Ω Step 2: supply currentI = V/Rtotal = 9.0 V / 6.0 Ω = 1.5 A All of this current passes through R1, which is in series with the battery.Step 3: p.d.sV1 = IR1 = 1.5 × 2.0 = 3.0 V The p.d.s around the loop must add up to the e.m.f., so the parallel pair has 9.0 − 3.0 = 6.0 V across it. R2 and R3 both have 6.0 V across them.Step 4: branch currentsI2 = 6.0/6.0 = 1.0 A and I3 = 6.0/12 = 0.50 A CheckAt the junction, 1.0 A + 0.50 A = 1.5 A, the supply current, so charge is conserved. The smaller resistor carries the larger current, as expected, and the p.d.s 3.0 V + 6.0 V add up to the 9.0 V e.m.f., so energy is conserved.