Week 12 · Chapter 4 Alternating Current Circuits (A): the sine waveform, frequency (50 Hz in Egypt) and the advantages of AC, why a moving-coil ammeter fails and the hot-wire ammeter, voltage and current in a resistor (in phase) and in a coil (voltage leads by 90°), inductive reactance XL = 2πfL, coils in series and parallel · 3 lessons
2
Voltage and current in a resistor and in a coil
Physics Sec 3 Term 1, Chapter 4: pp. 97–102
GoalSay how the voltage and the current are related in a resistor and in a pure coil, and use the graph to explain why the voltage leads by 90°.
New words
in phase · reaching zero and the peaks at the same momentspure coil · a coil whose own wire resistance is so small that we take it as zerovoltage leads · the voltage reaches its peak before the current does, here by a quarter of a cycle (90°)
Explanation
In a resistor, V = IR at every instant, so the voltage and the current rise, fall and change sign together: they are in phase. A pure coil is different. Its self-induced emf opposes every change of the current, so the source must supply a voltage V = L × ΔI ÷ Δt, which is proportional to the slope of the current curve. The slope of the sine curve is largest where it crosses zero and zero at its peak. So the voltage is at its peak when the current passes zero, and the voltage is zero when the current is at its peak. On the graph the voltage curve is a quarter of a cycle ahead of the current curve: the voltage leads the current by 90° (equally, the current lags the voltage by 90°). This holds at any frequency. In this course the current is taken as the reference and we say “leads” or “lags” for the voltage.
Worked examples
A 55 Ω resistor is connected to an AC supply of effective value 220 V. Ieff = V ÷ R = 220 ÷ 55 = 4 A, and the current is in phase with the voltage. The maxima are Vmax = 220 ÷ 0.707 = 311 V and Imax = 4 ÷ 0.707 = 5.66 A, and both are reached at the same moment.
Moment
Current I
Voltage V
0
0, rising
+Vmax
T ÷ 4
+Imax
0
T ÷ 2
0, falling
−Vmax
3T ÷ 4
−Imax
0
In a pure coil let I = Imax sin ωt. The voltage follows the slope, so it is a cosine curve. The table gives four moments in one cycle.
At 50 Hz, T = 20 ms, so 90° is a quarter of a cycle, 20 ÷ 4 = 5 ms. In a pure coil on the 50 Hz supply the voltage reaches each peak 5 ms before the current does.
Class activity
Paper only. Copy the table of Example 2 and draw the current as a sine curve over one cycle. Under it draw the voltage by reading the slope: steepest rising at the start, zero at the peak, steepest falling at half a cycle. Then check that your voltage curve is the current curve moved a quarter of a cycle to the left. Do the same for a resistor and say what is different.
Practice
1
A 100 Ω resistor is on an AC supply of effective value 200 V. Find Ieff and say what the phase difference is.
Ieff = 200 ÷ 100 = 2 A. In a resistor V and I are in phase, so the phase difference is 0°.
2
In a pure coil on the 50 Hz supply the voltage peaks at t = 2 ms. At what time does the current reach its peak (the next one)?
The voltage leads by a quarter cycle, 5 ms, so the current peaks at 2 + 5 = 7 ms.
3
In a pure coil the current passes through zero and is rising. Is the voltage zero or at its peak at that moment, and with which sign?
The slope of the current is largest and positive, so the voltage is at its positive peak, +Vmax.
4
Complete: in a resistor the phase difference is __°; in a pure coil the voltage ______ the current by __°.
