# Power Dissipated in a Sliding Resistor

## 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.

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gneill
Mentor
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 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.

• 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. ; _ ;

Last edited:
haruspex
Homework Helper
Gold Member
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
Mentor
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?

•  Thanks again. Your help is very much appreciated.

gneill
Mentor
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?

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

• 