Power Dissipated in a Sliding Resistor

[Silverado]

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.

[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.

 

[gneill]

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.

In the picture above the terminal c corresponds to the sliding tap connection. To carve the total resistance R_o 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.

 

[Silverado]

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 + R_aI₂ = 0
Lower loop: -R_aI₁ - R_aI₂ + E = 0

I substituted 100x for R_a and 100(1-x) for R_b. 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. ; _ ;

 

[haruspex]

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.

 

[gneill]

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?

 

[Silverado]

 

[Silverado]

Thanks again. Your help is very much appreciated.

 

[gneill]

Upper loop: -RI₃ + R_aI₂ = 0
Lower loop: -R_aI₁ - R_aI₂ + E = 0

Okay, so it's just a typo in the second loop equation. The first term of the lower loop should be -R_bI₁, since only I₁ flows through it according to your current definitions.

Can you show more of the steps you've taken to find I₃?

 

[Silverado]

 

[gneill]

Okay, looks good so far. You should be able to reduce further.

 

[Silverado]

Ty for your help & I am glad to be a part of the community! :-)