Week 14 · Chapter 4 (C): the series RC and LCR circuits (Z = √(R² + (XL − XC)²), the phase angle and the three cases, power only in the resistor), the oscillating LC circuit and f = 1/2π√(LC), resonance and the tuned circuit of a radio; the summary and exercises (pp. 114–116) are not used · 3 lessons
1
The series RC and LCR circuits
Physics Sec 3 Term 1, Chapter 4: pp. 107–113
GoalFind Z, the current, the phase angle and the power of series RC and LCR circuits, and decide whether the voltage leads or lags.
New words
net reactance · X = XL − XC: the coil and the capacitor work against each otherinductive / capacitive circuit · one where XL > XC (voltage leads) / XL < XC (voltage lags)power P · the average power used, P = I²R: only the resistor uses power
Explanation
In a series RC circuit the same current flows through R and C. VR is in phase with I, and VC lags I by 90°, so V = √(VR² + VC²), and Z = √(R² + XC²). The supply voltage lags the current, with tan θ = XC ÷ R. When a coil, a capacitor and a resistor are in series (LCR), VL leads by 90° and VC lags by 90°, so they point in opposite ways and partly cancel. Their difference is the net reactance X = XL − XC. Then Z = √(R² + (XL − XC)²) and tan θ = (XL − XC) ÷ R. If XL > XC the circuit is inductive and the voltage leads the current; if XL < XC it is capacitive and the voltage lags; if XL = XC it behaves like a resistor alone, Z = R. Coils and capacitors take energy and give it back in each cycle, so they use no power. The power used is P = I²R. In this course θ is the angle by which the supply voltage is ahead of the current (positive when XL > XC).
Net reactance
Name of the circuit
Voltage and current
Impedance
XL > XC
inductive
voltage leads the current
√(R² + (XL − XC)²)
XL < XC
capacitive
voltage lags the current
√(R² + (XL − XC)²)
XL = XC
behaves like a resistor
in phase
Z = R
Worked examples
Series RC: R = 90 Ω and XC = 120 Ω on a 300 V supply. Z = √(90² + 120²) = 150 Ω and I = 300 ÷ 150 = 2 A. VR = 180 V and VC = 240 V, and √(180² + 240²) = 300 V ✓. tan θ = 120 ÷ 90 = 4/3, so θ = 53° and the voltage lags the current.
Series LCR: R = 24 Ω, XL = 50 Ω, XC = 18 Ω on a 120 V supply. Net X = 50 − 18 = 32 Ω. Z = √(24² + 32²) = 40 Ω and I = 120 ÷ 40 = 3 A. P = I²R = 3² × 24 = 216 W. tan θ = 32 ÷ 24 = 4/3, θ = 53°, and XL > XC, so the circuit is inductive. Check: VR = 72 V, VL = 150 V, VC = 54 V, and √(72² + (150 − 54)²) = 120 V ✓.
When XL = XC. R = 15 Ω, XL = XC = 40 Ω on 90 V. Z = R = 15 Ω, I = 90 ÷ 15 = 6 A, P = 6² × 15 = 540 W. The coil and the capacitor each have 6 × 40 = 240 V across them, much more than the 90 V supply, but the two voltages are opposite and cancel in the circuit.
Class activity
Paper only. For R = 8 Ω fill a table with three rows: (XL, XC) = (10, 4), (4, 10) and (7, 7) Ω. For each find the net X, Z, whether the circuit is inductive, capacitive or resistive, and whether the voltage leads, lags or is in phase. Then say in one sentence why a coil and a capacitor in series can make the current larger than either one alone would allow.
Practice
1
A series RC circuit has R = 30 Ω and XC = 40 Ω on 100 V. Find Z and I, and say whether the voltage leads or lags.
Z = √(30² + 40²) = 50 Ω. I = 100 ÷ 50 = 2 A. A capacitor is the reactive part, so the voltage lags the current.
2
A series LCR circuit has R = 16 Ω, XL = 10 Ω and XC = 22 Ω on 60 V. Find X, Z, I, P and say whether it is inductive or capacitive.
X = XL − XC = 10 − 22 = −12 Ω. Z = √(16² + 12²) = 20 Ω. I = 60 ÷ 20 = 3 A. P = 3² × 16 = 144 W. XL < XC, so it is capacitive and the voltage lags (tan θ = −12 ÷ 16 = −0.75, θ = −37°).
3
R = 8 Ω, XL = 20 Ω and XC = 14 Ω are in series on 120 V. Find Z, I and the power used.
X = 20 − 14 = 6 Ω. Z = √(8² + 6²) = 10 Ω. I = 120 ÷ 10 = 12 A. P = 12² × 8 = 1152 W.
4
Why does a pure coil or a pure capacitor use no power on average, although current flows through it?
It takes energy from the source in one quarter of the cycle and gives it back in the next (as magnetic or electric energy), so nothing is turned into heat.
