1) An efficient way of making a light dimmer is to place a variable inductor in

Question

1) An efficient way of making a light dimmer is to place a variable inductor in series with a light bulb. We’ll think of the light bulb as an ideal resistor with R=100?. The inductor has no resistance and L can be varied from 0 to Lmax. The electrical supply is 120V (rms) at 60Hz.

a) What setting for L will maximize the power dissipated in the light bulb?

b) What is this maximum power?

c) The minimum power is 10% of the maximum power. What is Lmax?

d) The inductor could be replaced by a resistor to achieve the same reduction in the power

dissipated in the light bulb. What resistor would be required?

e) What is the advantage of using an inductor instead of the resistor?

Explanation / Answer

Vrms = 120 V, R = 100 ohms, w = 60 Hz

a) When L = 0, XL = wL = 0, Requivalent = R + 0 = R = 100

b) Hence imax = 120/100 = 1.2 A (rms) => Pmax = imax2*Requivalent = 1.22*100 = 144 Watts

c) Pmin = 0.1 Pmax = 14.4 W => Pmin = Vrms2/Requivalent => 14.4 = 1202/Requivalent => Requivalent = 1000

1000 = Requivalent = R + XL = 100 +XL => XL = 900 = wL = 60*L => L = 15 Henry (this is Lmax as XL is max)

d) Then the new Rnew = XL (max) = 900 ohms (as calculated above)

e) beacuse varying the value of L (inductance) in an inductor is easy, so then we can vary the overall resistance and hence the brightness of the bulb, while using resisitor will require to change the resistor each and every time to change the power dissipated at the bulb

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