الأسبوع 13 · الفصل الرابع (٢): المكثف وشحنه وتفريغه مع التيار المستمر والمتردد، والمفاعلة السعوية وتوصيل المكثفات، والمعاوقة، ودائرة (R–L) على التوالي · 3 دروس
3
Impedance and the series RL circuit
Physics Sec 3 Term 1, Chapter 4: pp. 102–107
GoalFind the impedance Z and the current of a series RL circuit, explain why the voltages add as vectors, and find the phase angle.
New words
impedance Z · the total opposition of a circuit to AC, in ohms: V ÷ I with effective valuesphase angle θ · the angle by which the supply voltage is ahead of (or behind) the currentvector sum · adding quantities that point at right angles by Pythagoras, not by simple addition
Explanation
When a resistor R and a coil XL are in series, the same current I flows through both. The voltage across the resistor, VR = IR, is in phase with I. The voltage across the coil, VL = IXL, leads I by 90°. These two voltages reach their peaks at different moments, so they do not add up by simple addition: they are at 90° to each other and the supply voltage is their vector sum, V = √(VR² + VL²). Dividing by I gives the impedance, Z = √(R² + XL²), in ohms, and the current is I = V ÷ Z (effective values). The supply voltage leads the current by the phase angle θ, with tan θ = VL ÷ VR = XL ÷ R. The picture is a right-angled triangle with R across, XL up and Z as the long side. XL = 2πfL is worked out with π = 3.14, as in Week 12.
Worked examples
Part
Value
Z
100 Ω
I
2 A
VR
120 V
VL
160 V
VR + VL (wrong)
280 V
√(VR² + VL²)
200 V
A series circuit has R = 60 Ω and XL = 80 Ω on a 200 V supply. Z = √(60² + 80²) = √(3600 + 6400) = 100 Ω and I = 200 ÷ 100 = 2 A. VR = 2 × 60 = 120 V and VL = 2 × 80 = 160 V. Their plain sum is 280 V, more than the 200 V supply, but their vector sum is √(120² + 160²) = 200 V. tan θ = 80 ÷ 60 = 4/3, so θ = 53° and the voltage leads the current.
R = 50 Ω in series with a coil of L = 0.2 H on a 160 V, 50 Hz supply. XL = 2 × 3.14 × 50 × 0.2 = 62.8 Ω. Z = √(50² + 62.8²) = √6443.8 = 80.3 Ω. I = 160 ÷ 80.3 = 1.99 A. VR = 1.99 × 50 = 99.5 V and VL = 1.99 × 62.8 = 125 V. tan θ = 62.8 ÷ 50 = 1.256, so θ is about 51.5°, the voltage leading.
Spot the error. A student measures VR = 30 V and VL = 40 V in a series RL circuit and says that the supply is 70 V. The two voltages are 90° apart, so V = √(30² + 40²) = √2500 = 50 V.
Class activity
Paper only. On squared paper draw a right-angled triangle with the horizontal side R = 9 Ω and the vertical side XL = 12 Ω (1 cm for 3 Ω). Measure the long side with a ruler, convert it to ohms, and compare it with √(9² + 12²). Then measure the angle with a protractor and compare it with the value from tan θ = 12 ÷ 9.
Practice
1
A coil with XL = 12 Ω is in series with R = 9 Ω. Find Z.
Z = √(9² + 12²) = √225 = 15 Ω.
2
The circuit of the last question is on a 45 V supply. Find I, VR and VL, and check that VR and VL give 45 V.
I = 45 ÷ 15 = 3 A. VR = 3 × 9 = 27 V. VL = 3 × 12 = 36 V. √(27² + 36²) = √2025 = 45 V ✓.
3
In a series RL circuit VR = 24 V and VL = 70 V. Find the supply voltage.
V = √(24² + 70²) = √5476 = 74 V (not 94 V).
4
R = 16 Ω and XL = 12 Ω are in series. Find Z, tan θ and θ (use the 3-4-5 triangle), and say whether the voltage leads or lags.
Z = √(16² + 12²) = 20 Ω. tan θ = 12 ÷ 16 = 0.75, so θ = 37°. A coil is in the circuit, so the voltage leads the current.
