Thanks Jason for all your hard work. I've applied all of your equations and I still don't get the answer the book gives. I don't know how you got the answer but I've come to the conclusion that the answer the book gives is wrong. Here is my reasoning. First, the power rating is only a rating, not power dissipated. Second, in order to solve for voltage you must know current or power dissipated and to solve for current you must know the voltage or power dissipated. None of those variables are given in the word problem. The only variables given are the power rating and resistance. Therefore the only equation that will work is V max which is the square root of the power rating times the resistance. This is the only logical solution.
I agree that the power rating is only a rating – but I still think that it can be used to solve the problem - which I think is solvable. I think that each power rating can be used to determine a highest common value of current for each resistor – which will be one of two values that I think can be solved for. Using, P = (I^2)*R, for one of the resistors, the power rating may actually be equal to power dissipated. This is because we are looking for a highest safest value this is limited by one of the resistors. I think that even this lower value might be considered dangerous – and is why the power rating of resistors may in practice be increased.
I don’t agree that to find the voltage one has to know the current and power dissipated – if this is what you meant. I’d consider V = IR. Can I be solved for?
I don’t agree that to find the current one has to know voltage and power dissipated – if this is what you meant. According to P = (I^2)*R, only P and R have to be known. P was discussed above. The value of R is given for each resistor. This will result in two currents. Which one is safe? Maybe the lowest. Why?
After having solved for the current I, V = I*R can be used.
I think that this is right. The only way that I can think of it being wrong is if P doesn’t equal (I^2)*(R1 + R2) even though it does equal (I^2)*(RT) – using the same symbols described in an earlier post.
I tried to explain in my last post why I didn’t think that the formula that you gave would work. I’ll try to simplify my explanation – not because I am sure that I am right, but because I don’t think that I explained it very well. The power rating in the equation I think actually means two different things – the power dissipated and the power rating. These may be the same for one resistor but not the other.
Basically, there is a circuit. We can apply any voltage to it. According to that voltage, there will be a certain current that flows through the circuit. That current – together with the resistors, will determined power dissipated - and perhaps damage the circuit. Which voltage is best? We don’t want to go too high or low when applying voltage. So, we consider the poProxy-Connection: keep-alive
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the resistors – maybe becauseProxy-Connection: keep-alive
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e don’t want to do any damage. So, how much current? That depends on P and R – which Proxy-Connection: keep-alive
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e given. They are given for each resistor. We can solve for each resistor and come up with two answers for the current I that will work for each resistor. Will they work for the other resistor as well? Maybe only if only the lower one is used – because power dissipated increases as current increases. Does this make sense according to P = (I^2)*R? So, a lowest current can be selected. Then the formula V = I(R + R) can give the voltage. I think that all this depends on if P = (I^2)*(R1 + R2).
Basically, there is a circuit. We can apply any voltage to it. According to that voltage, there will be a certain current that flows through the circuit. That current – together with the resistors, will determined power dissipated - and perhaps damage the circuit.. Which voltage is best? We don’t want to go too high or low when applying voltage. So, we consider the power rating of the resistors – maybe because we don’t want to do any damage. So, how much current? That depends on P and R – which are given. They are given for each resistor. We can solve for each resistor and come up with two answers for the current I that will work for each resistor. Will they work for the other resistor as well? Maybe only if only the lower one is used – because power dissipated increases as current increases. Does this make sense according to P = (I^2)*R? So, a lowest current can be selected. Then the formula V = I(R + R) can give the voltage. I think that all this depends on if P = (I^2)*(R1 + R2).
The reason why I am thinking at this time that it might work is because P = IV, V = IR, and RTotal = R1 + R2.