Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    3
    ReputationRep:
    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.
    Offline

    20
    ReputationRep:
    (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
    • Thread Starter
    Offline

    3
    ReputationRep:
    Thank you so much, I understand now
    (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
    • Study Helper
    Offline

    21
    ReputationRep:
    Study Helper
    (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:

    \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}}
 
 
 
Poll
Do you agree with the PM's proposal to cut tuition fees for some courses?

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.