Week 13 · Chapter 4 (B): the capacitor (Q = CV) on DC and on AC (the current leads, the voltage lags by 90°), capacitive reactance XC = 1/2πfC, capacitors in series and in parallel, impedance and the series RL circuit (Z = √(R² + XL²), phase angle, voltages add as vectors) · 3 lessons
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Capacitive reactance, capacitors in series and in parallel
Physics Sec 3 Term 1, Chapter 4: pp. 102–107
GoalCalculate XC = 1 ÷ (2πfC) and the current in a capacitor, and find the total capacitance of capacitors in series and in parallel.
New words
capacitive reactance · the opposition of a capacitor to an alternating current, XC = 1 ÷ (2πfC), in ohmscapacitors in parallel · they have the same voltage, and the capacitances add: C = C₁ + C₂ + …capacitors in series · they carry the same charge, and 1 ÷ C = 1 ÷ C₁ + 1 ÷ C₂ + …
Explanation
A capacitor opposes an alternating current by its capacitive reactance XC = 1 ÷ (2πfC), in ohms when f is in hertz and C in farad. We use π = 3.14 and change µF to F first. A higher frequency or a larger capacitance moves the charge more easily, so XC is smaller. At f = 0 (DC) XC grows without limit and no steady current passes. With effective values I = V ÷ XC, and the current leads the voltage by 90°. Capacitors joined in parallel have the same voltage, and their charges add, so C = C₁ + C₂ + …. Capacitors joined in series carry the same charge, and their voltages add, so 1 ÷ C = 1 ÷ C₁ + 1 ÷ C₂ + …, which makes the total smaller than the smallest capacitor. This is the opposite of resistors. Reactances behave like resistances: in series XC = XC1 + XC2 + …, which agrees with the rule for C.
f (Hz)
25
50
100
200
XC of 50 µF (Ω)
127.4
63.7
31.8
15.9
Worked examples
A capacitor of 50 µF is on a 100 V, 50 Hz supply. XC = 1 ÷ (2 × 3.14 × 50 × 50 × 10⁻⁶) = 1 ÷ 0.0157 = 63.7 Ω. I = V ÷ XC = 100 ÷ 63.7 = 1.57 A.
Connection
Same quantity
Total C
parallel
voltage
50 µF
series
charge
12 µF
Capacitors of 20 µF and 30 µF. In parallel C = 20 + 30 = 50 µF. In series C = (20 × 30) ÷ (20 + 30) = 600 ÷ 50 = 12 µF, smaller than 20 µF.
For the same 50 µF capacitor, doubling the frequency halves the reactance: 63.7 Ω at 50 Hz, 31.8 Ω at 100 Hz, 15.9 Ω at 200 Hz. The graph of XC against f falls and never reaches zero; towards f = 0 it climbs without limit.
Class activity
Paper only. Make a table of XC for a capacitor of 20 µF at 10, 50, 100 and 500 Hz (π = 3.14). By what factor does XC change when the frequency is multiplied by 10? Then say which kind of signal, low or high frequency, passes more easily through a capacitor.
