How to solve direct and indirect/inverse variation problem?

….just talk your way thru this…...

1. wire2 is 3 x longer than wire1….... thus resistance of wire2 has to be 3 x larger than wire1 on the basis of length, alone.

2. wire 2 has a diameter that is 1.5 x larger than wire1…. thus resistance of wire2 has to be 1.5^^2 = 1.5*1.5 times smaller than wire 1 on the basis of diameter , alone.

3. the cumulative effect is that wire 2 is 3/1.5*1.5 = 2/1.5 = 1.333… times larger than wire 1….. or in other words 33%larger than wire 1.

In equations one has R1 =k * L1/D1^^2
R2 =k * 3*L1/(1.5*D1)^^2

…... you can solve it from there…... <g>

…..

Specifically what step do you not understand?

@malcolm.knapp I guess I really don’t understand where to begin solving.

@malcolm.knapp ...you want to solve for R2 relative to R1…... i.e. R2/R1 = ??

…my bad…..... that should have been addressed to @scotielee

@virtualist I guess that I am just having trouble setting up the problem. I don’t have too much of an issue setting up problems that are ONLY direct or ONLY inverse, but when they’re both, it confuses me in structuring a formula.

Ok. All it’s telling you is that in general

resistance=(some number)x(length)/((diameter)^2)

“Varies directly with length” means that as the length increases the resistance does, too. “Varies inversely with the square of the diameter” means that as the diameter squared gets larger the resistance gets smaller.

You’ll have one of these equations for each wire. The “some number” will be the same for each wire if they are each made out of the same material (which you have to assume in the problem as you stated it), so when you’re comparing these two equations you can cancel out that “some number”.

Think of it as two problems in sequence. First you use the values from the first wire to find the resistance/cm and then you use that value to calculate the resistance of the second wire.