9.6.2 Example Series Resonant
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A circuit contains a 10 Ω resistor, 0.25 H inductor, and 40.52 μF capacitor connected in series to a 240 V, 50 Hz power supply. The current flowing is calculated to be 24 A, with 240 V dropped across the resistor, and 1885 V dropped across both the inductor and capacitor due to their equal reactances of 78.54 Ω at 50 Hz.
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9.6.2 Example Series Resonant
- 1. Example A 10 Ω resistor, 0.25 H inductor and a 40.52 µF capacitor are connected in series across a 240 V 50 Hz supply. Calculate the current flowing and voltage drop across each component.
- 3. XL = 2 π f L XL = 2 x π x 50 x 0.25 XL = 78.54 Ω
- 4. XC = 1 . 2πfC XC = 1 2 x π x 50 x (40.52 x 10¯⁶) = 1 0.01273 = 78.54 Ω
- 5. Z = √ R² + (XL — Xc)² Z = √ 10² + (78.54 — 78.54)² Z = 10 Ω
- 6. I = V Z I = 240 = 24 V 10
- 7. VR = I.Z = 24 x 10 = 240V VL = I.XL = 24 x 78.54 = 1885 V VC = I.Xc = 24 X 78.54 = 1885 V