Question

A Light-Dimmer Circuit The intensity of a lightbulb with a resistance of 160 Ω is controlled by connecting it in series with an inductor whose inductance can be varied from L=0 to L=Lmax. This "light dimmer" circuit is connected to an ac generator with a frequency of 60.0 Hz and an rms voltage of 120 V .
The inductor is now adjusted so that L=Lmax. In this case, the average power dissipated in the lightbulb is one-fourth the value found in part A. What is the value of Lmax?
ANSWER IN UNITS OF H.

Answer 1

$Solution:\\A)\;P=\frac{U^2}{R}=\frac{120^2}{160}=90W;\\B)\;\frac{1}{4}P=I^2R; P=4I^2R;I=\frac{U}{Z}\\P=\frac{4U^2}{Z^2}R; Z=2U\sqrt{\frac{R}{P}}=240\times\sqrt{\frac{160}{90}}=320 \Omega;\\Z=\sqrt{R^2+X_{L_{max}}^2};\\X_{L_{max}}=\sqrt{Z^2-R^2}=277.13\Omega;\\X_{L_{max}}=2\pi fL_{max};\\L_{max}=\frac{X_{L_{max}}}{2\pi f}=\frac{277.13}{6.28\times60}=0.735H;\\Answer:L_{max}=0.735H$