Question

please help for part B

A Light-Dimmer Circuit The intensity of a lighbub with a resistance of 100 is controlled by connecting in series with an inductor whose inductance can be varied from 0 to L-This sight immer circuit is connected to an ac generator with a frequency of 600 Hz and an mms votage of 120 Part A What is the average power dissipated in the lightbulb when L-02 Express your answer using two significant figures P-90 W Previous new Completed Part The inductor is now adjusted so that L... In this case, the average power dissipated in the lightbulb in ore-bout the value found in part A What is the value of L? Express your answer using two significant figures ΑΣ και ? H Submit Best An

Answer 1

Part A) The power dissipated in lightbulb,

P = 90 W

Part B) The value of inductance

$\small L_{max} =0.73 H$

Explanation for part B :-

Part 6) Now the indretance of the inductor, L = L max Therefore the ampendance, 2 - JR+ w Londok where, angular forquency raf Herre given that, = 120N me voltage, Vorms R 160 R Resistance frequensey, f = 6000 Hz And average power dresipated an bulb from part A P = gow (Op 2

P = Vreme Vrms ( 2 ) P = Von R here Poneers in this case er i th of the power when L=o. 90 w XR 4 Vine (R2 + x)) Noms & R R²+ (W Lmaan 9o 2 W²=40&fe 120² x 160 90 11 160² + (HTX60² Cuar 22.5 120²x160 1602 +48 24602xLmas)

160² + bon? 60%* Emax 2 120 X 160 22.5 160² + 472 60² Lmax 102400 4T²60² Lomax = 102 400-1602 76800 2 L max 76800 and 49²60² Limax = 0.5403 Lmax 0.5403 L max = 0.735 H

Hence the answer