00:01
Hi there, so for this problem, we are told that a resistor with a resistance capital r is connected to a battery with an f that is given and that nph is equal to 16 volts.
00:17
We're also given an internal resistance for this, that is 0 .370 oms.
00:25
And finally, we are given a power dissipated of 81 watts.
00:37
So the question is, for what two values of the resistance will be the power dissipated in this resistor? okay? so we know that the current in the circuit depends on this resistance capital r and on the internal resistance of the battery.
00:58
As well as the nph of the battery.
01:02
It is only the current in the resistance that dissipates energy in the resistance.
01:09
So we will have that the current is given then by the nph, divided by the resistance plus the internal resistance lower case are.
01:21
And we know that the power is defined as the current to the square times the resistance.
01:27
So what we are going to do now is to simply substitute the current into the power.
01:38
So the power now becomes the nph to the square divided by the resistance plus the internal resistance and that to the square.
01:49
And this times the resistance capital r.
01:53
So in here, if we solve and we start solving, we'll obtain that the nph to the square...