The diagram below shows a block (made of material of ?o = 0.6 ?·m, ? = 0.004 K-1
ID: 1443361 • Letter: T
Question
The diagram below shows a block (made of material of ?o = 0.6 ?·m, ? = 0.004 K-1) with dimensions ? = 8 cm, w = 1 cm, and h = 4 cm. It is maintained (by external means) at room temperature (20 Celsius).Two very large plates (not shown) which are maintained at a fixed voltage difference of 5 V are variously applied to the block as indicated below.
a.) Find the current that flows through the block if the voltage difference is applied:
across the left and right faces
I = ___ mA
across the top and bottom faces
I = ___ mA
across the front and back faces
I = ___ mA
b.) Suppose that the block's temperature was instead maintained (by external means) at 165 oC.
The current that would flow through the block in each case is smaller than what you found in part a, by a factor of _____ times.
Here are the correct answers, but I have trouble understanding a few parts of this solution.
The unit in question is alpha = 0.004K-1
In part B, the person used (4/1000). But doesn't K-1 mean the reciprocal, which would make it 1/(4/1000) instead?
Explanation / Answer
a) Resistance = resistivity x Length / cross section area
R = rho L / A
and voltage V is applied across the block then
current I = V / R
when left to right face.
L = l = 8 cm = 0.08 m
A = w x h = 0.01 x 0.04 = 4 x 10^-4 m^2
R = (0.6 x 0.08) / (4 x 10^-4) = 120 ohm
I = V/R = 5 / 120 = 0.04167 A = 41.67 mA
b) when top to bottm face.
L = h = 4 cm = 0.04 m
A = w x l = 0.01 x 0.08 = 8 x 10^-4 m^2
R = (0.6 x 0.04) / (8 x 10^-4) = 30 ohm
I = V/R = 5 / 30 = 0.1667 A = 167.67 mA
c) when front to back face.
L = w = 4 cm = 0.01 m
A = h x l = 0.04 x 0.08 = 32 x 10^-4 m^2
R = (0.6 x 0.01) / (32 x 10^-4) = 1.875 ohm
I = V/R = 5 / 1.875 = 2.6667 A = 2667.67 mA
b) as temparture increases, Resistance also increase with temperature.
new R, R' = R [ 1 + alpha(deltaT)]
here alpha = 0.004 1/K
deltaT = Tf -Ti = (165 + 273)K - (273 + 20)K = 145 K
R' = R [ 1 + (0.004 1/K)(145K)]
R' = R [ 1 + (0.004 x 145) (K/K) ] = R [ 1 + 0.58 ] = 1.58R
so new curret, I' = V / R' = V / (1.58R ) = 0.633 V/R
initially current was I = V/R
I' = 0.633V/R = 0.633I
so new current I' is 0.633 times initial current I.