- #1

Lambda96

- 157

- 59

- 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##.

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##.