Today at work I was thinking hypothetically about a simple circuit that I was unable to get an answer that made any sense. What if you had a simple 12V source with a 5Ω resistor in series, also in series with a black box that consumes 60W of power. You aren't given any characteristics about the black box other than it dissipates 60W. When I do a KVL around the simple loop, I obtained 12V - 5i - 60/i = 0, since P = VI and for this device V and I are both unknown. If I multiply through by i, you can obtain a quadratic formula, but neither of the results physically make sense; in this example they result in complex numbers, and even the magnitude of the real part isn't near what I was expecting. This was driving me nuts today and it is so simple, but somewhere I am not right. Any ideas?
The maximum power available from a 12V battery through a 5 ohm resistor will be when the resistor drops half the voltage, then current is 6V / 5R = 1.2 amp. Wmax to the load is therefore 1.2A * 6V = 7.2 watt. How can you get 60 watt from the black box if it does not have an alternative supply of energy? Is it consuming negative energy = generating energy.
True story. The black box has an external power supply....maybe like an op amp or something of that nature. So if you are confused how the 12 volt battery gets that much power to black box....don't be, because it can't with the 5 ohm resistor dicating the story.
Indeed as Baluncore said: solution of your 12i-5i^2+60=0 it is i=[12+/-sqrt(144-4*5*W)]/2/5. This will be a complex number while the black box power is more than 7.2 w[144/4/5]. So 7.2 W it is maximum allowable power for this black box.
I agree with you, sophiecentaur. However, since OP said “I was thinking hypothetically “, we may continue hypothetically.:shy: Let's say the black box ratings are: Prated=60 W; Vrated=12 V. Then Rrated=12^2/60=2.4 ohm Now i=12/(5+2.4)=1.62 A. The power dissipated by black box Pact=2.4*1.62^2=6.3 W If the black box temperature will be now lower than rated then the black box resistance will be lower. For 30oC difference the copper will be (1+0.00393*30)=1.12 times less. Ract=2.4/1.12=2.147 ohm. i=12/(5+2.147)=1.68 A; Pact=2.147*1.68^2=6.06 W[!!!].
Many threads get launched by a newcomer and then PF 'regulars' get into it and give it 'the treatment' (I include myself here). Sometimes, I think we often read too much into the OP and all that was really necessary may have been to give some clarification - or to point out a possible mistake. Our enthusiasm can scare people off and we don't want that. I was just hoping for a bit of feedback before too much of an avalanche.