
#1
Jan810, 01:54 PM

P: 408

1. The problem statement, all variables and given/known data
Alright I was given an assignment and I have done all the questions, but I'm not sure if they are right and would like some help if they aren't. 1. A wire has resistivity 5.6 x 10^8 Ωm at 20 °C and a temperature coefficient of resistance 4.5 x 10^3 °C1^1. What length of the wire (radius 12.5mm) would you need to make a heating element that would dissipate 1000 watts? The heating element operates at 220 volts and has a temperature of 700°C. 2.________________40Ω _40v80Ω l l  (ignore the ______ that's the only way it would let me place the 40ohm and 10ohm there. ___l _____________10Ω___l  So there is a voltage source of 40v connected in series with an 80 ohm resistor connected in series with two resistors (40 ohm and 10 ohm) that are connected in parallel with eachother. It asks to find the current, the power dissipated by each of the 3 resistors, and the power supplied by the 40v source. 3. We have a sawtooth wave, current with respect to time. The current increases linearly from 0 seconds to 1 second with respect to time with a slope 1 so the current starts at 0 and ends at 1A, at 1 second it drops to a current of 1 and increases for one more second with a slope 1 until the current is at 0. We also have a power source with a 5000Ω resistor. Calculate the total energy supplied by the source to the resistance over the 2 second period. 4. The last question gives us a graph of current vs. voltage (current on the yaxis, voltage on the xaxis), the initial current is 3A and initial voltage is 0V, it decreases linearly from 01V with a slope of 3. So initial current and voltage is 3A and 0V, and final current and voltage is 0A and 1V. Find the open current voltage of the cell, the short circuit current, and the maximum power that the cell can deliver to an external resistance. 3. The attempt at a solution 1. I=P/E=1000W/220V=50A/11 R=V/I = 220V/(5A/11)=484Ω ρ=ρ0(1+αΔT) = (5.6 x 10^8Ωm)(1+4.5 x 10^3 °C1^1(700°C20°C)) = 2.27x10^7 Ωm. L=RA/ρ = [484x((2.5x10^4)/2)^2]/2.27x10^7 Ωm = 1.0 x 10^2m Can someone tell me whether what I did is correct or not? 2. a) To find the current I found the equivalent resistance of the resistors in parallel, = (1/10+1/40^)^1=8Ω, then I find the total resistance which is 80Ω+8Ω=88Ω. I then did I=V/R = 40v/88Ω= 5A/11. b)For the power dissipated I used the equation P=I^2R. For R1, P=(5A/11)^2(80Ω)=2000W/121 For R2 (10Ω) I found the current to be 4A/11, and P=(4A/11)^2(10Ω)=160W/121 For R3 (40Ω), P=(1A/11)^2(40Ω)=40W/ c) For the power supplied by the 40v source, I did P=IE=(5A/11)(40v)=200W/11 3. What I did was integrated the current with respect to time to get the total current supplied, so the equation of the line from 01 second is y'=xdx, and the integral is (x^2)/2. However at 1 second the current drops from 1 to 1 and increases for one second until zero, so the area under the graph from 0 to 1 second is the same as the area under the graph from 12 seconds, so I added the areas together: (x^2)/2+(x^2)/2=x^2. I integrated it from 01 second to get 1A. P=I^2R=(1A)^2(5000Ω)=5000W E=PT=5000W(2seconds)=10 000J Edit: Ok I think for this one I wasn't allowed to just add the areas together, I had to integrate each function separately, correct? (can someone briefly explain why I can't just add the areas of the triangles together though?) 4. a) For the open circuit voltage of the cell I simply looked at when the current was 0 on the graph, so the answer is 1V. b) For the shortcircuit current I looked when the voltage was 0, so the answer is 3A. c) For the maximum power that the cell can deliver to an external resistance I did I=3V+3 (from the graph) and multiplied that by V because P=IV =(3V+3)V=3V^2+3V. I then derived this equation to find when the power is at a maximum, so dP=6V+3 which equals 0 at V=1/2. So the maximum power that the cell can deliver to an external resistance is P=3(1/4)+3(1/2) = 3/4W So I did my best to do these problems, I don't have anyone I'm close with in the class to ask questions too especially since the class just started so help is much appreciated! Thanks. 



#2
Jan910, 01:06 AM

HW Helper
P: 2,324

All of your answers look fine except #3. The question says that I=t, so the integral of power (P=I^2*R) with respect to time should be the integral of t^2*R dt from 0 to 1, not the integral of R*tdt from 0 to 1.
You can add the two areas together; don't worry about that. It's clear that the two areas represent the same power usage. 



#3
Jan910, 10:56 AM

P: 408

But what I did was integrated current with respect to time, and then plugged it into the equation P=I^2R.....so the total area under the triangle is the total current, and then I plugged in the total current into that equation, why doesn't that work?




#4
Jan910, 05:42 PM

P: 210

CIRCUITS Assignment Help Please :)
A few things:
#1  The wire is circular, so A is pi*r^2, not just r^2. Also, R should be 48.4, not 484 (because I = 50/11, not 5/11). I get 10.45m as my final answer. Other than that, I get the same answers. The reason you can't add up the area under the current for #3 is that you need the integral of I^2, not I (I think). Edit: #3: I is in mA on the graph: the answer should now be 3.3x10^3 J. 



#5
Jan910, 05:59 PM

P: 408

Ah thanks for telling me about the 48.4, stupid mistake. And yeah for the area I did it with pi included but I forgot to write it.
Yeah you're right about the I^2, but you can add the I^2's together, all you have to do is 2x(integral from 0 to 1) of Rt^2dt. Ends up being the same thing. Anyway thanks again. 



#6
Jan1010, 06:02 PM

P: 452

I think I am in your class. In problem 3 isn't the current measured in mA?




#7
Jan1010, 06:04 PM

P: 408

Yeah it is, I forgot about that. But all I had to do was divide the answer by 1 000 000.




#8
Mar311, 08:23 PM

P: 1

Thanks for the tips. I'll try to answer no.3.




#9
Mar311, 09:44 PM

Mentor
P: 11,417

In #1, you've calculated a resistivity of 2.27x10^{7} Ωm for operation at °C. If the wire radius is r = 12.5mm, then the wire is only 4.63 x 10^{4} Ω/m (given by ρ/Area).
If the required resistance is 48.4Ω, then you'll need over 104 km of wire! Are you sure about the wire radius figure? 


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