Physics cylindrical conductor

A cylindrical conductor of length, l diameter D and resistivity p has resistance R.
What is the resistance of another cylindrical conductor of length l, diameter D/2 and resistivity p.

Can anyone explain the calculations?

I know that the resistance would be 4R but can't figure out the workings.

Reply 1

Joinedup

20

Original post by Paypurr

A cylindrical conductor of length, l diameter D and resistivity p has resistance R.
What is the resistance of another cylindrical conductor of length l, diameter D/2 and resistivity p.

Can anyone explain the calculations?

I know that the resistance would be 4R but can't figure out the workings.

The formula book gives you...
ρ=RA/L

ρ is resistivity (it's a greek letter rho fwiw)
R is Resistance
A is cross sectional area
L is length

this can be rearranged to make R the subject

R=ρL/A

the relationship between area of a cylinder (circular cross sectional area) and the diameter is that area is proportional to the square of the diameter... so if you half the diameter, the area is reduced to (1/2)² i.e. 1/4

since A is on the bottom of the fraction in R=ρL/A that has the effect of multiplying the resistance by 4

Reply 2

username2676349

OP

17

Thank you so much, I understand now :smile:

Original post by Joinedup

The formula book gives you...
ρ=RA/L

ρ is resistivity (it's a greek letter rho fwiw)
R is Resistance
A is cross sectional area
L is length

this can be rearranged to make R the subject

R=ρL/A

the relationship between area of a cylinder (circular cross sectional area) and the diameter is that area is proportional to the square of the diameter... so if you half the diameter, the area is reduced to (1/2)² i.e. 1/4

since A is on the bottom of the fraction in R=ρL/A that has the effect of multiplying the resistance by 4

Reply 3

uberteknik

21

Original post by Paypurr

A cylindrical conductor of length, l diameter D and resistivity p has resistance R.
What is the resistance of another cylindrical conductor of length l, diameter D/2 and resistivity p.

Can anyone explain the calculations?

I know that the resistance would be 4R but can't figure out the workings.

Proof:

Do this as a ratio:

\frac{R_{D}}{R_{\frac{D}{2}}}

As Joinedup said

\rho = \frac{RA}{L}

rearranging for R

R = \frac{\rho L}{A}

then substitute for the two surface areas :

Be careful here,

A_{D} = \pi r^{2} = \pi (\frac{D}{2})^{2} = \pi(\frac{D^{2}}{4})

and

A_{\frac{D}{2}} = \pi r^{2} = \pi (\frac{\frac{D}{2}}{2})^{2} = \pi(\frac{D}{4})^{2} = \pi(\frac{D^{2}}{16})

substituting the respective areas:

Unparseable latex formula:

\frac{R_{D}}{R_{\frac{D}{2}}} = \frac{\left{[}\frac{\rho L}{\pi(\frac{D^{2}}{4})}\right{]}}{\left{[}\frac{\rho L}{\pi(\frac{D^{2}}{16})}\right{]}}

Cancelling out terms appearing in both the numerator and denominator on the RHS leaves:

\frac{R_{D}}{R_{\frac{D}{2}}} = \frac{4}{16} = \frac{1}{4}

and finally

4R_{D} = R_{\frac{D}{2}}

Physics cylindrical conductor

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