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
3
Inductive reactance, coils in series and in parallel
Physics Sec 3 Term 1, Chapter 4: pp. 97–102
GoalCalculate the inductive reactance XL = 2πfL and the current in a coil, and combine coils in series and in parallel.
New words
inductive reactance · the opposition of a coil to an alternating current, XL = 2πfL, in ohmsself-inductance L · the property of a coil in henry (H) that sets the emf it induces in itselfeffective values · the values a meter shows; the rule I = V ÷ XL uses them
Explanation
The faster the current changes, the larger the self-induced emf of the coil, and the more the coil opposes the current. A larger L opposes more too. The opposition of a pure coil to an AC is its inductive reactance XL = 2πfL, in ohms when f is in hertz and L in henry. We use π = 3.14. With effective values the current is I = V ÷ XL, and the voltage leads the current by 90°. At f = 0 (a steady current) the reactance is zero, so a pure coil lets a DC pass freely. Because XL is proportional to f, its graph against frequency is a straight line through the origin. Convert mH to H before you start (1 mH = 10⁻³ H). For coils far apart, so that one does not act on the other, the reactances add in series, XL = XL1 + XL2 + …, and in parallel 1 ÷ XL = 1 ÷ XL1 + 1 ÷ XL2 + …, exactly like resistors.
f (Hz)
25
50
100
200
XL of 0.5 H (Ω)
78.5
157
314
628
Worked examples
A pure coil of L = 0.5 H is on a 110 V supply of 50 Hz. XL = 2 × 3.14 × 50 × 0.5 = 157 Ω and I = 110 ÷ 157 = 0.70 A. At 100 Hz, XL = 314 Ω and I = 110 ÷ 314 = 0.35 A: twice the frequency, half the current.
A coil of 20 mH on the 50 Hz supply. First 20 mH = 0.02 H. XL = 2 × 3.14 × 50 × 0.02 = 6.28 Ω.
Connection
Rule
XL
series
XL1 + XL2
150 Ω
parallel
XL1·XL2 ÷ (XL1 + XL2)
33.3 Ω
Two coils have XL = 100 Ω and 50 Ω at the same frequency. In series XL = 100 + 50 = 150 Ω. In parallel XL = (100 × 50) ÷ (100 + 50) = 33.3 Ω, smaller than the smaller coil.
Class activity
Paper only. Make a table of XL for a coil of L = 0.25 H at 10, 50, 100 and 500 Hz, and plot the points on squared paper (use π = 3.14). What kind of line do they lie on, and what does it tell you about XL and f? What is XL at f = 0?
Practice
1
Find XL of a coil of L = 0.2 H at 50 Hz (π = 3.14).
XL = 2πfL = 2 × 3.14 × 50 × 0.2 = 62.8 Ω.
2
Find XL of the same coil at 150 Hz.
XL = 2 × 3.14 × 150 × 0.2 = 188.4 Ω (three times the frequency, three times XL).
3
A pure coil of 80 mH is on a 40 V supply of 50 Hz. Find XL and the current.
L = 0.08 H. XL = 2 × 3.14 × 50 × 0.08 = 25.12 Ω. I = 40 ÷ 25.12 = 1.59 A.
4
Two coils have XL = 60 Ω and 40 Ω. Find the total XL in series and in parallel.
