00:01
Hello students, for part a we want to find the power dissipated by the bulb.
00:08
From figure 1 we can see that motor is initially not connected.
00:13
So we can write i into r plus 0 .4 which is equal to 120.
00:22
That is i equal to 120 divided by r plus 0 .4.
00:28
Now it is given that power dissipated across bulb which is equal to 75 watts.
00:37
That can be written as i square r which is equal to 75.
00:43
That is 130 divided by r plus 0 .4 whole square which is equal to 75.
00:52
This is not 130 this is 120 which is equal to 120 square equal to 75 into r plus 0 .4 whole square.
01:05
Now from this we can find r.
01:10
That is r equal to 199 .19 ohm and 8 .36 into 10 to the power minus 3.
01:21
Here this value is very small.
01:24
So we can take r equal to 191 .19 ohm.
01:30
Now this is the resistance of the bulb...