The Choke, the Wattmeter and the 4 µF Capacitor
A shopkeeper's 40 W fluorescent batten is rated for 220 V, 50 Hz and uses a magnetic choke in series with the tube. He complains that it flickers at start-up. A technician measures the rms current drawn from the mains as 0.41 A, while a wattmeter in the same circuit reads 36 W. The supply is a clean 220 V sinusoid. When the tube is struck it sustains about 90 V rms across itself. The shopkeeper has been told the choke “wastes current” and proposes to replace it with a wire-wound resistor of the same resistance, which is cheaper, keeping the existing starter. He also has a 4 µF, 400 V capacitor in his spares box and asks whether connecting it across the mains at the batten would reduce his electricity bill. He pays for real energy only, at a flat rate per kilowatt-hour.
1. From the ammeter and wattmeter readings, calculate the apparent power and the power factor. Explain where the “missing” power goes, and check whether the choke behaves as a nearly pure inductance.
Read the worked answer
Apparent power. S = V_rms I_rms = 220 × 0.41 = 90.2 VA. Real power. P = 36 W (wattmeter). Power factor. cos φ = P/S = 36/90.2 = 0.40 (φ ≈ 66°). Reactive power. Q = √(S² − P²) = √(90.2² − 36²) ≈ 82.5 VAr.
Where the “missing” 54 W goes. Nowhere - it is not consumed. In an inductive circuit the current lags the voltage by φ; the instantaneous power is positive for part of the cycle (energy stored in the choke's magnetic field) and negative for the rest (returned to the supply). The wattmeter, which averages over a cycle, records only the net unidirectional flow, 36 W. The 82.5 VAr shuttles back and forth.
Is the choke nearly pure? Real power in the tube at strike ≈ V_tube I = 90 × 0.41 = 36.9 W, which accounts for essentially all of the wattmeter reading. A pure inductance in series would give V_choke = √(220² − 90²) ≈ 201 V, and V_choke I = 201 × 0.41 ≈ 82 VA, matching the reactive power above. So the choke is very nearly ideal; its copper loss is only a watt or two. The choke does two jobs: it limits the tube current after strike, and with the starter it produces the inductive kick (L di/dt) that strikes the tube.
2. The shopkeeper replaces the choke with a resistor chosen so that the current is still 0.41 A. Calculate the resistance and the power wasted in it, and explain why the lamp may not start at all.
Read the worked answer
Total impedance needed. |Z| = 220/0.41 = 537 Ω. The struck tube passes 0.41 A at 90 V, so it behaves as R_tube ≈ 90/0.41 = 220 Ω. The series resistor must therefore be about 537 − 220 = 317 Ω.
Power wasted. P_R = I²R = (0.41)² × 317 ≈ 53 W - dissipated as heat in the resistor, more than the 36 W the tube itself uses. Compare the choke, whose copper loss is a few watts because the energy is stored in a field and returned each cycle. The “cheaper” resistor roughly doubles the running cost and heats the fitting.
Why the lamp may not strike. A fluorescent tube is a gas discharge with negative resistance: before strike it needs a voltage far above the running value. The starter heats its bimetallic contacts, opens the circuit, and the choke's sudden collapse of current produces a large L di/dt pulse that strikes the tube. A resistor produces no such inductive kick - the current simply stops, with no voltage spike - so the tube may glow at the ends and never start, or start erratically. The resistor must also be rated well above 53 W and mounted where it cannot scorch the fitting.
3. Evaluate the 4 µF capacitor proposal: calculate the capacitor current, the new line current and the new power factor, and state clearly what the shopkeeper does and does not save.
Read the worked answer
Capacitor current. X_C = 1/(2πfC) = 1/(2π × 50 × 4 × 10⁻⁶) ≈ 796 Ω, so I_C = 220/796 ≈ 0.276 A, leading the voltage by 90°.
Resolve the lamp–choke current (0.41 A at cos φ = 0.40):
- in phase (active): I_p = 0.41 × 0.40 = 0.164 A
- lagging (reactive): I_q = 0.41 × 0.916 ≈ 0.376 A
After connecting the capacitor the reactive component partially cancels: net I_q = 0.376 − 0.276 = 0.100 A. New line current = √(0.164² + 0.100²) ≈ 0.19 A; new power factor = 0.164/0.19 ≈ 0.85.
What is saved. The wattmeter still reads about 36 W, and the capacitor consumes no real power, so the energy charge on the bill is unchanged - the shopkeeper saves nothing on his kWh. What falls is the current drawn from the supply (0.41 A → 0.19 A), which reduces I²R losses in the wiring and the supply transformer and frees capacity for other consumers; this is why utilities, not consumers, care about power factor. The capacitor must be rated for mains use (400 V a.c. is adequate, and a motor-run type is preferred) and must be switched with the lamp, since with the lamp off it would draw 0.276 A of purely reactive current.