Term 1 review: circuits, magnetism, induction and AC
Answers are for parents and teachers.
1
Two-loop circuit laid out as in Week 3: branch 1 has E₁ = 9 V (+ pole up) and R₁ = 2 Ω, branch 2 has E₂ = 4 V (+ pole up) and R₂ = 1 Ω, the middle branch has R₃ = 5 Ω. Assume I₁ and I₂ upward and I₃ downward. Find the three currents, say which battery is being charged, and find the voltage across R₃.
I₃ = I₁ + I₂. Loop A: 9 = 2I₁ + 5I₃ = 7I₁ + 5I₂. Loop B: 4 = I₂ + 5I₃ = 5I₁ + 6I₂. Multiply A by 6 and B by 5: 54 = 42I₁ + 30I₂ and 20 = 25I₁ + 30I₂. Subtract: 34 = 17I₁, so I₁ = 2 A. Then 5I₂ = 9 − 14, so I₂ = −1 A and I₃ = 1 A. Check in B: 5 × 2 + 6 × (−1) = 4 ✓. I₂ is negative, so battery 2 is being charged. The voltage across R₃ is 5 × 1 = 5 V.
2
A coil of N = 80 turns measures 0.10 m × 0.06 m. (a) As a motor coil, it carries 1.5 A in a field of 0.25 T with its plane parallel to the field. Find the torque. (b) Turned at 50 Hz in the same field as a generator (π = 3.14), find emf(max) and the effective value.
A = 0.10 × 0.06 = 6 × 10⁻³ m². (a) τ = BIAN = 0.25 × 1.5 × 6 × 10⁻³ × 80 = 0.18 N·m. (b) ω = 314 rad/s, emf(max) = NBAω = 80 × 0.25 × 6 × 10⁻³ × 314 = 37.7 V. Effective value = 0.707 × 37.7 = 26.6 V.
3
A station sends 60 kW along a line of resistance 2 Ω. Find the current and the power lost in the line at 3 kV and at 12 kV, and by what factor the loss falls.
At 3 kV: I = 60 000 ÷ 3000 = 20 A and loss = 20² × 2 = 800 W. At 12 kV: I = 5 A and loss = 5² × 2 = 50 W. The voltage is four times larger, the current four times smaller, and the loss 800 ÷ 50 = 16 times smaller.
4
R = 20 Ω, L = 0.1 H and C = 50 µF are in series on a 76 V, 50 Hz supply (π = 3.14). Find XL, XC, Z, I and the power used, and say whether the voltage leads or lags the current.
XL = 2 × 3.14 × 50 × 0.1 = 31.4 Ω. XC = 1 ÷ (2 × 3.14 × 50 × 50 × 10⁻⁶) = 63.7 Ω. X = 31.4 − 63.7 = −32.3 Ω. Z = √(20² + 32.3²) = 38.0 Ω. I = 76 ÷ 38.0 = 2.0 A. P = I²R = 4.0 × 20 = 80 W. X is negative, so the circuit is capacitive and the voltage lags the current (tan θ = −1.6, about −58°).
5
Spot the error. An ohmmeter has a total internal resistance of 4000 Ω. Its pointer rests at one seventh of full scale. A student says: “Rx is seven times 4000 Ω, that is 28 000 Ω.” Find the mistake and the correct Rx.
At one seventh of full scale the current is Ig ÷ 7, so the total resistance in the circuit is seven times Ri: 7 × 4000 = 28 000 Ω. That total includes the instrument’s own 4000 Ω, so Rx = 28 000 − 4000 = 24 000 Ω (six times Ri).