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Power Dissipated in a Sliding Resistor

  • Thread starter Silverado
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Homework Statement



The circuit [http://tinypic.com/r/2cwl7bt/9] shows a battery connected across a uniform resistor R0. A sliding contact can move across the resistor from x = 0 at the left to x = 10 cm at the right. Find an expression for the power dissipated in the resistor R as a function of x. Plot the function for E = 50 V, R = 2000 Ω, and R0 = 100 Ω.

upload_2016-3-20_13-40-36.png

[Note: picture embedded by moderator]

Homework Equations


[/B]
P = V2R = i2R = iV

P = energy/time

The Attempt at a Solution



I really do not know where to begin. My professor did not explain how to approach problems involving sliding contacts. I tried to use the second equation because it involved the variable time, which I figured I could somehow relate to the distance x. Any help would be appreciated.
 

Answers and Replies

  • #2
gneill
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Hi Silverado, Welcome to Physics Forums.

Please make sure that any images crucial to your problem statement are visible in your post. Helpers should not have to follow links to off-site images in order to understand the question. Thanks.

A resistor with a sliding contact (also called a potentiometer) can be treated as two separate resistors. The sliding contact is associated with the junction point between the resistors. The sum of the two resistances is a constant value equal to the stated value of the potentiometer. In your case that would be 100 Ω.

upload_2016-3-20_13-53-45.png


In the picture above the terminal c corresponds to the sliding tap connection. To carve the total resistance Ro into two pieces it's convenient to use a parameter that divides the total resistance into two parts. Thus you might make x the parameter and:

##R_a = x⋅R_o##
##R_b = (1 - x)⋅R_o##
where 0 ≤ x ≤ 1.

See what you can do with that information to make an attempt at the problem.
 
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  • #3
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I tried the problem again using your advice and I think I am doing too much algebra. I used your info. in addition to the loop and junction rule. I reconstructed the circuit with two loops, one containing resistors R and Ra and the other Ra and Rb.

JR:
I1 = I2 + I3

LR:

Upper loop: -RI3 + RaI2 = 0
Lower loop: -RaI1 - RaI2 + E = 0

I substituted 100x for Ra and 100(1-x) for Rb. I then proceeded to solve for I3. The algebra was very intense and I think I am doing something wrong. The resulting I3 came out to be x/(-2x^2 - 2x + 40). Please, any input would help. This problem makes me sad. ; _ ;
 
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haruspex
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The algebra was very intense and I think I am doing something wrong. The resulting I3 came out to be x/(-2x^2 - 2x + 40). Please, any input would help. This problem makes me sad. ; _ ;
Almost right. You have a sign error. You can check whether it is right by trying x=0 and x=1. x=1 gives the wrong result.
Not sure why you found the algebra so hard. I can't tell whether there is an easier way without seeing all your working.
 
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  • #5
gneill
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There will be a fair bit of algebra involved, but nothing too onerous.

It's not clear how you've defined your currents, but the same total current should flow through a given resistor in both of your loop equations, so something's amiss with those equations. Can you describe (or better, show with a picture) the currents that you're using?
 
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Thanks again. Your help is very much appreciated.
 
  • #8
gneill
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Upper loop: -RI3 + RaI2 = 0
Lower loop: -RaI1 - RaI2 + E = 0
Okay, so it's just a typo in the second loop equation. The first term of the lower loop should be -RbI1, since only I1 flows through it according to your current definitions.

Can you show more of the steps you've taken to find I3?
 
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  • #10
gneill
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Okay, looks good so far. You should be able to reduce further.
 
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  • #11
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Ty for your help & I am glad to be a part of the community! :-)
 

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