A light dimmer is a possible application of a series RL-circuit. Assume for simplicity that the resistance of the lightbulb is constant with R=100 Ohm. For a dimmer, the bulb is connected in series with a variable inductance L.
The circuit is connected to a wall outlet with 115V AC oscillating at 60 Hz. What range of L is needed if it is desired
to vary average power from 30W to 100W?
2007-11-22 12:00:28 · 2 answers · asked by Ali Ahmed 2 in Science & Mathematics ➔ Engineering
First calculate the two resistances required to get 30 and 100W. You can transpose E=IR and P=EI into V^2/W=R. I get 441 and 132 ohms.
So now you need XL's of 441 - 100 = 341 ohms and 132 - 100 = 32 ohms.
Tranpose XL = 2(PI)fL into XL/(2(PI)f) = L to solve for your two L's. Those are the upper and lower values of inductance needed.
2007-11-23 02:04:47 · answer #1 · answered by Gary H 6 · 0⤊ 1⤋
1. Use P = I^2 R to calculate the current for 30 and 100 watts.
2. Use Z = V/I to calculate the 2 values of circuit impedance.
3. Calculate the impedances of the inductor that must be in series with the fixed 100 Ohms.
4. Use L = Z/Omega to get the answer.
2007-11-22 23:22:42 · answer #2 · answered by Tim C 7 · 0⤊ 0⤋