Calculate Max Resistance for 2 Parallel Wires

• Lambda96
In summary, the conversation discusses the calculation of the total resistance of two resistors connected in parallel. The participants also consider the length of two wires made of the same material and use a derivative to find the maximum resistance. They also discuss the importance of verifying whether the extremum found is a maximum or a minimum. Finally, they mention the equivalence of finding the extremum of the inverse of the total resistance and finding the maximum of the equivalent resistance.
Lambda96
Homework Statement
How must the lengths ##L_1## and ##L2## be distributed so that the resistance between the points ##X## and ##Y## becomes maximum
Relevant Equations
none
Hi,

I am not sure if I have calculated the task here correctly:

Based on the drawing, I now assumed that the two resistors are connected in parallel. The total resistance can then be calculated as follows ##\frac{1}{R_T}=\frac{1}{R_1}+\frac{1}{R_2}##.

Since the two wires are made of the same material, I only considered their length in the calculation. Then I specified the following ##L_1=x## and ##L_2=1000-x## and substituted into the equation above ##\frac{1}{x}+\frac{1}{1000-x}##

Since the resistance is supposed to be maximum, I calculated the derivative of the above expression and then set it to zero.

$$\frac{1}{(1000-x)^2}-\frac{1}{x^2}=0$$

After that, I simply solved the equation for x and got for ##x=500m## which gives me the following for the lengths ##L_1=500m## and ##L_2=500m##.

Looks fine, but you should verify you found a maximum rather than a minimum.

Lambda96 and erobz
Lambda96 said:
Then I specified the following ##L_1=x## and ##L_2=1000-x## and substituted into the equation above ##\frac{1}{x}+\frac{1}{1000-x}##
I think you have treated the above expression as the total resistance. But it is not - it is the inverse of the total resistance.

Also, a good way of checking answers (when practical) is to compare the answer against extreme cases. E.g. what resistance would you expect if ##L_1=L_2##? What resistance would you expect if ##L_1= 1000m## and ##L_2=0##?

Lambda96
Steve4Physics said:
I think you have treated the above expression as the total resistance. But it is not - it is the inverse of the total resistance.
Of course, since ##f(x) = \frac{1}{x}## is strictly monotone decreasing (barring the boundary where ##x = 0##), searching for an extremum on ##\frac{1}{x}## is largely equivalent to searching for an extremum on ##x##.

Lambda96
jbriggs444 said:
Of course, since ##f(x) = \frac{1}{x}## is strictly monotone decreasing (barring the boundary where ##x = 0##), searching for an extremum on ##\frac{1}{x}## is largely equivalent to searching for an extremum on ##x##.
Yes. Finding the extremum of ##\frac 1{R_T}## (which is a minmum here) is equivalent to finding the maximum of ##R_T##.

But it's not clear (to me anyway) if that's what the OP has intentionally done!

Lambda96 and erobz
Isn't it easier to find an expression for the equivalent resistance and just look at it?
$$\frac{1}{R_{eq}}=\kappa\left(\frac{1}{L-x}+\frac{1}{x}\right)=\frac{\kappa~L}{x(L-x)}\implies R_{eq}=\frac{1}{\kappa}x(L-x).$$ This is equivalent to the problem of having a string of length ##L## and being asked to find the rectangle of perimeter ##L## that has maximum area. The answer is a rectangle of equal sides because for every rectangle of base ##b## and height ##h## there is another rectangle of base ##h## and height ##b## that has equal area. Thus, any ##b\neq h## is not unique and cannot be an extremum. Only ##b=h## gives a unique area and hence an extremum. It is a maximum because when either one of the sides goes to zero, the area goes to zero.

Last edited:
Lambda96 and nasu
vela said:
Looks fine, but you should verify you found a maximum rather than a minimum.
If you are referring to the expression that OP is optimizing, it better have a minimum because it is the inverse of the equivalent resistance.

Lambda96
Thank you vela, Steve4Physics, jbriggs444 and kuruman, for your help and for checking my calculation

berkeman

1. What is the formula for calculating the maximum resistance for 2 parallel wires?

The formula for calculating the maximum resistance for 2 parallel wires is R = (R1 * R2) / (R1 + R2), where R1 and R2 are the individual resistances of the wires.

2. How do I determine the individual resistances of the wires?

The individual resistances of the wires can be determined by using a multimeter to measure the resistance of each wire separately.

3. Can I use this formula for any type of wire?

Yes, this formula can be used for any type of wire as long as the resistances are known and the wires are connected in parallel.

4. What is the maximum resistance if the two wires have the same resistance?

If the two wires have the same resistance, the maximum resistance will be half of the individual resistance of each wire. This can be calculated using the formula R = R1 / 2.

5. Is there a limit to the number of parallel wires that can be used in this calculation?

No, there is no limit to the number of parallel wires that can be used in this calculation. The formula can be applied to any number of parallel wires as long as the individual resistances are known.

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