Answer to Question #3827 in Electric Circuits for mia
<br>a. What is the effective voltage?
<br>b. A 60 W light bulb is placed across the generator. A maximum current of 0.70 A flows
through the bulb. What effective current flows through the bulb?
<br>c. What is the resistance of the light bulb when it is working?
b. Effective current is Imax/√2= 0.7/√2 = 0.49 I
c. The resistance is R = 2W/Imax2 from the law W = I2/R = Imax2 /2R
R = 2*60 / 0.72 = 244.89 Ohm
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!