How to do these two questions? Can give me some hints? Thank you In 2.51 How can we setting up R is infinite.. R is infinite-->power is infinite(T-To is infinite)? Actually I have no idea of how to set up the equations..As the value of T is unknown.. In 2.65 By KVL VG-V conductor+VM=0 if VM=105V V conductor=215V if VM=117V V conductor=227V I think that the range of resistance can be found And then what should I do?
Does anyone interested in these questions? I have thought it for a long period of time I would like to have clues of thinking..
kit.kit44, All right, I will try to do my best on the first problem. We start with P=(I^2)*R . The problem says that the fuse blows when R becomes infinite. From the above equation, that implies that the fuse blows when an infinite amount of power is dissipated in the fuse resistance if the current is at its rated valuer. So... P = (I^2)*R ===> I^2 = P/R ===> P/[0.11(1+0.7(0.35*P)] Dividing P/[0.11(1+0.7(0.35*P)] by P gives 1/[(0.11/P)+0.11*0.7*0.35] As P becomes infinite 1/[(0.11/P)+0.11*0.7*0.35] becomes 1/(0.11*0.7*0.35) = 37.11 So I^2 = 37.11 and I = 6.09 amps Ratch
I get the same answer, but I am confused: How can a resistor dissipate power when it is approaching infinite resistance, unless it is driven by a current source with infinite voltage compliance? If the voltage is limited, then the current must approach zero as the resistance approaches infinity. it seems to me that the phase change of the metal needs to be taken into account, but this leads to a discontinuous function of resistance vs current. I suppose that in reality, the power dissipation only has to get high enough to melt the fuse, and the equation is only an approximation.
Ron H Here are some resistance and current values for different power dissipations. The resistances were calculated from the formula given. 100 watts, 2.805 ohms, 5.971 amps 1000 watts, 27.060 ohms, 6.079 amps 1000000 watts, 26950 ohms, 6.0914 amps 1000000000000 watts, 2.695E10 ohms, 6.0914 amps As you can see, if the fuse could dissipate those ridiculous power values, the current hardly changes. Since we don't have the textbook, and the problem does not tell us, we don't know at what temperature the wire in the fuse will melt and how to derate the fuse. If we did know the melt temperature T, then we could figure from T-To = k*P what the power dissipation of the fuse is, but its current rating would still be around 6 amps before derating it. So as far as I can see, we did as much calculation as we could with the information we were given. Ratch
Thanks for doing those calculations, Ratch. 100 watts, 2.805 ohms, 5.971 amps, 16.75 volts 1000 watts, 27.060 ohms, 6.079 amps, 165.4 volts 1000000 watts, 26950 ohms, 6.0914 amps, 164,160 volts 1000000000000 watts, 2.695E10 ohms, 6.0914 amps, 1.6416E11 volts. My point was that the voltage compliance has to be infinite to actually achieve infinite power dissipation (infinite resistance). Otherwise, if the fuse didn't melt before it reached infinite resistance, it would never fail. However, I suppose ≈6 amps is close enough for government work.
Its not a very sensible question. The fuse melts when the temperature exceeds the melting point. They are also obliquely making the point that the actual melting point is not very important as the current is strongly limited by the dissipated heat causing the resistance to increase indefinitely (in theory). Actually the fuse has to both melt and the molten fusewire has to fall away from its conections. By that time the power dissipation may have vaporised the liquid metal