Alternating current, resistors and coils
Answers are for parents and teachers.
1
An alternating current has a period of 4 ms. Find its frequency, and the time of a quarter cycle.
f = 1 ÷ 0.004 = 250 Hz. A quarter cycle is 4 ÷ 4 = 1 ms.
2
A 25 Ω resistor is on an AC supply of effective value 150 V. Find Ieff, Vmax and Imax.
Ieff = 150 ÷ 25 = 6 A. Vmax = 150 ÷ 0.707 = 212 V. Imax = 6 ÷ 0.707 = 8.49 A.
3
A pure coil of L = 0.8 H is on a 200 V supply (π = 3.14). Find XL and the effective current at 50 Hz and at 100 Hz.
At 50 Hz: XL = 2 × 3.14 × 50 × 0.8 = 251.2 Ω, I = 200 ÷ 251.2 = 0.80 A. At 100 Hz: XL = 502.4 Ω, I = 200 ÷ 502.4 = 0.40 A.
4
A pure coil has XL = 125.6 Ω on the 50 Hz supply (π = 3.14). Find its inductance L.
L = XL ÷ (2πf) = 125.6 ÷ (2 × 3.14 × 50) = 125.6 ÷ 314 = 0.4 H.
5
For a pure coil on the 50 Hz supply, say which of voltage and current is ahead, by what angle and by how many milliseconds, and say why the voltage is largest when the current passes through zero.
The voltage is ahead by 90°, which is a quarter of 20 ms, that is 5 ms. The voltage needed is V = L × ΔI ÷ Δt, the slope of the current curve, and the slope is largest where the current crosses zero.