00:01
In this problem, we have been given the following circuit and we are required to find first the voltage that's applied in the circuit.
00:10
So for that, let's compute the total resistance first.
00:15
So that's the second part and we first compute that second part which will help us get the total voltage that's applied here.
00:23
So the total resistance that will be found when we add the resistors individually in pairs.
00:29
So we see that these two resistors, they're in series.
00:33
So their total resistance will be 16 -oam.
00:35
And this 16 -on resistor and this 64 -on resistor, they are in parallel.
00:41
So we just use this expression to compute their equivalent resistance, which will be 64 times 16 over 64 plus 16.
00:49
So when we multiply 64 with 16 and divide this with 64 plus 16, we're going to get the resistance coming out to be 12 .8.
00:59
Oms and this resistor is now in series with the 7 .2 ome resistor so their equivalent resistance will be 12 .8 plus 7 .2 om and that's 20 om and this 20 ome resistor it's in parallel with this 30 ome resistor so their equivalent resistance it will be 30 times 20 over 30 plus 20 so when we multiply 30 with 20 and divided with 30 plus 20 we can get the total resistance coming out to be 12 ome.
01:33
And as we observe that, the current supplied by the source is 5 ampiers and the total resistance is 12 oom.
01:41
So we just use ome slot to compute the voltage.
01:45
So according to omsdlotte, voltage is current times resistance and that's 60 volt, which is the answer here.
01:52
And in the next case, we have to determine the power that's delivered by the source.
01:58
So to compute the power, we just use the expression, we times i.
02:04
So the voltage is 60 and the current is 5.
02:07
So the power is 300 watts.
02:11
And we have to now find out the power which is dissipated in this 10 -on resistor...