00:01
In this question we have a voltage source of 120 volts.
00:09
We have the wire resistance being indicated as an additional resistor, but it's to represent the wire resistance.
00:22
We have two objects that are actually connected in parallel.
00:27
That is the bulb, the light bulb, as well as the motor.
00:38
Now at the start, given that the motor is not when it's turned off, when it's just the bulb.
00:48
The bulb will operate normally at 75 watts of power and the voltage is 120 volts.
00:57
Since the resistance is very, very small, 0 .8 oms, we can assume that most of the emf will go to the bulb.
01:08
Now we know that p is equals to v square over r.
01:15
So if you have to assume that the voltage across our bulb would be 120 volts at normal operation, then the resistance would be v square over p.
01:30
We take 120 square divided by the power and we get the resistance of the bulb to be 192...