The problems are on the kirchhoff's rules worksheet. Try them on your own first, then compare with the answers below.
Quick answer: In a DC circuit, a wire's resistance is R = ρL/A, series resistors add, and parallel resistors share the same voltage. These worked answers correct the 20 problems of the AP Physics 2 worksheet on kirchhoff's rules, from lamp resistance to dimming lamp: each solution shows the formula, the substitution and the final answer with its unit.
Each answer shows the formula, the substitution and the final result in bold. Compare your method as well as your numbers, because AP scoring rewards a clear line of reasoning even when the arithmetic slips.
Level: AP Physics 2 (algebra-based) · Curriculum: AP P2 Unit 11
Problem 1 Lamp resistance
1. R = V/I = 4.5 / 0.30 = 15 Ω.
2. P = IV = 0.30 × 4.5 = 1.4 W (1.35 W).
Problem 2 Wire dimensions
1. Ω·m (ohm metre).
2. R is proportional to L, so doubling the length doubles R.
3. A is proportional to d², so doubling d makes A four times larger and R four times smaller.
Problem 3 Three in series
1. Req = 10 + 22 + 33 = 65 Ω.
2. I = V/R = 13 / 65 = 0.20 A.
Problem 4 Two in parallel
1. 1/R = 1/20 + 1/30 = 5/60, so R = 12 Ω.
2. I = 6.0 / 12 = 0.50 A (0.30 A + 0.20 A through the two branches).
Problem 5 Heating element
1. 0.20 mm² = 0.20 × 10⁻⁶ m² = 2.0 × 10⁻⁷ m².
2. R = ρL/A = (1.1 × 10⁻⁶)(2.0) / (2.0 × 10⁻⁷) = 11 Ω.
3. I = 5.5 / 11 = 0.50 A.
Problem 6 Junction currents
1. Current in = 3.0 + 2.0 = 5.0 A. Out: 1.2 + I = 5.0, so I = 3.8 A.
2. Conservation of charge.
Problem 7 Long wait
1. A fully charged capacitor blocks current: 0 A.
2. V = IR = 0 V.
3. The battery emf is shared only by the capacitor, so its voltage equals the battery emf.
Problem 8 Charge and energy
1. Q = CV = (220 × 10⁻⁶)(9.0) = 2.0 × 10⁻³ C (1.98 mC).
2. U = ½CV² = 0.5 × 220 × 10⁻⁶ × 81 = 8.9 × 10⁻³ J.
Problem 9 Series-parallel network
1. 1/Rp = 1/6.0 + 1/3.0 = 3/6.0, so Rp = 2.0 Ω. Req = 8.0 + 2.0 = 10 Ω.
2. I = 15 / 10 = 1.5 A.
3. V across the pair = 1.5 × 2.0 = 3.0 V, so I = 3.0 / 6.0 = 0.50 A.
Problem 10 Series or parallel lamps
1. Series: R = 24 Ω, I = 0.25 A, P = IV = 0.25 × 6.0 = 1.5 W.
2. Parallel: R = 6.0 Ω, I = 1.0 A, P = 6.0 × 1.0 = 6.0 W.
3. Each lamp gets 6.0 V in parallel (3.0 W each) but only 3.0 V in series (0.75 W each), so parallel is brighter.
Problem 11 Three bulbs
1. A carries the whole battery current I. B and C share it, so each carries I/2. Power is I²R, so A is brightest and B and C are equally dim.
2. With C removed, the total resistance rises from 1.5R to 2R. The current in A falls from ε/1.5R to ε/2R, so A dims. The current in B rises from ε/3R to ε/2R, because it no longer shares, so B brightens.
Problem 12 Length and resistance data
1. R/L = 4.87 / 0.80 = 6.1 Ω/m.
2. R/L = ρ/A, so ρ = (R/L)A = 6.09 × 8.0 × 10⁻⁸ = 4.9 × 10⁻⁷ Ω·m.
Problem 13 Aluminium cable
1. A = π(0.225 × 10⁻³)² = 1.59 × 10⁻⁷ m².
2. R = (2.8 × 10⁻⁸)(120) / (1.59 × 10⁻⁷) = 21 Ω.
3. I = 12 / 21.1 = 0.57 A.
Problem 14 Battery with internal resistance
1. I = ε/(R + r) = 9.0 / 4.5 = 2.0 A.
2. V = ε − Ir = 9.0 − 2.0 × 0.50 = 8.0 V.
3. P = I²r = 4.0 × 0.50 = 2.0 W.
Problem 15 Two batteries
1. I₃ = I₁ + I₂.
2. Loop 1: 9.0 − 1.0I₁ − 2.0I₃ = 0. Loop 2: 3.0 − 1.0I₂ − 2.0I₃ = 0.
3. Substituting I₃: 3I₁ + 2I₂ = 9.0 and 2I₁ + 3I₂ = 3.0. Solving gives I₁ = 4.2 A, I₂ = −1.8 A, I₃ = 2.4 A. The negative sign means the real current in the right branch flows downward through the 3.0 V battery, so that battery is being charged.
Problem 16 Designing a heater coil
1. A = π(0.20 × 10⁻³)² = 1.26 × 10⁻⁷ m².
2. L = RA/ρ = 24 × 1.26 × 10⁻⁷ / 1.1 × 10⁻⁶ = 2.7 m.
3. P = V²/R = 240² / 24 = 2400 W.
Problem 17 Capacitor in a network
1. At first the uncharged capacitor acts like a wire, so R₂ is in parallel with R₃: 300 × 100 / 400 = 75 Ω. Total 175 Ω, I = 20 / 175 = 0.11 A.
2. Voltage across the pair = 0.1143 × 75 = 8.57 V, so I₃ = 8.57 / 100 = 0.086 A.
3. After a long time the capacitor branch carries no current, so I = 20 / 400 = 0.050 A. V across R₂ = 15 V, and R₃ has no drop, so VC = 15 V. Q = CV = 10 × 10⁻⁶ × 15 = 1.5 × 10⁻⁴ C.
Problem 18 Reading a charging graph
1. The curve levels at 6.0 V, the battery emf.
2. 63 % of 6.0 V is 3.8 V, reached at about 20 s, which is τ.
3. R = τ/C = 20 / (1000 × 10⁻⁶) = 2.0 × 10⁴ Ω.
4. At t = 0 the capacitor acts like a wire: I = 6.0 / 20 000 = 3.0 × 10⁻⁴ A.
Problem 19 Spot the errors
1. The student added the resistor drops to the battery emf. Crossing a resistor along the current is a drop, so 12 − 2I − 4I = 0 and I = +2.0 A.
2. Resistors in parallel have a smaller total than either one: 1/R = 1/10 + 1/10, so R = 5.0 Ω. The student added them as if they were in series.
Problem 20 Dimming lamp
1. I = 6.0 / (5.0 + 1.0) = 1.0 A. P = I²R = 5.0 W.
2. Parallel lamps: 2.5 Ω. I = 6.0 / 3.5 = 1.71 A. Terminal voltage = 6.0 − 1.71 × 1.0 = 4.3 V. Each lamp: P = V²/R = 4.29² / 5.0 = 3.7 W.
3. The extra current causes a larger drop across the internal resistance, so the voltage across each lamp falls from 5.0 V to 4.3 V and the lamp dims.
