Problem 11: Acylindrical conductor has resistance of Ro 520 2. It has length l,
ID: 1527717 • Letter: P
Question
Problem 11: Acylindrical conductor has resistance of Ro 520 2. It has length l, radius of the cross section r, and resistivity g Otheexpertta.com Part (a) Express the resistance in terms of ,r, g. Expression Select from the variables below to write your expression. Note that all variables may not be required. a, B, H, g, e, d, g, h,i,j,l,m,P, r,t Part (b) If the resistivity increases by a factor of 4, what would the value of the new resistance be, in ohms? Numeric A numeric value is expected and not an expression. Part (c) If the length of the conductor decreased by a factor of what would the value of the new resistance be, in ohms? Numeric A numeric value is expected and not an expression. Part (d) If the radius of the conductors increased by a factor of 5, what would be the value of the new resistance, in ohms? Numeric A numeric value is expected and not an expression.Explanation / Answer
if resistivity is Q
a)
then resistance R = Q*l/(pi*r^2)
b)
if resistivty increase by a factor of 4 then
resistnce also increase by a factor of 4.
so new R = 4*R
c)
if the length of the conductor decreased by a factor of 4 then
resistance also decrease by a factor of 4.
new R = R/4
d)
if radius of the conductor decrease by a factor of 5 then
resistance will increase by a factor of 5^2 = 25.
new R = 25*R