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Unformatted text preview: 27.2.1 Concepts in Ohm': Law Given two conductors #1 and #2, they are of the same
material. For i=1, 2, denote ° length: Li, 0 cross sectional area: Ai, ° applied voltage Vi: ° field inside : Hi 0 average drift velocity: vdi. V!
Given: A2=Al, L2=2Ll, V1=V2=V.
Find: Ratios E2!E1_ 3
E2/El 2 1 1/2 Hint: V=EL. J=UE=nqu
Extra: Find delvdl. Explanation: EZJEI=(V2:’L2)I(V1:’L0:112.
Extra: Same ohmic materials imply that 01:62
and nlznz. Or
vdzivdl=J2IJl=(02E2)/(o1E1)=1/2. 27.5.1 Average collision time For an ohmic conductor, we have vdzat,a=£,andJ=nqu.
m So the drift speed or the current density is proportional to
E. This relation holds only if r is strictly constant. Estimate how much will 1: be changed, if vdrjﬁ is doubled,
for a typical case where the average thermal speed is vth~103km/s, and average drift speed, vdliﬁ~10-4m/s. 1. It increases by more than 0.01%.
It stays essentially the same, i.e.
the change is less than 0.01%. 3. It decreases by more than 0.01%. Hint: The average speed of free electrons |V|=IVth + Vdriﬁl. Explanation: Based on vector the reader should convince
himself/herself that the percentage of the maximum change
2v . —4
2X 10 _
dr'ﬂ =—6=2><10 8%
than 0.01% Ans=2. in v is less than . This is less 27.3.1 Thermal effect on resistivity Visualize free electrons moving through a crowded
medium. They collide with the atoms along the way. As
the temperature increases, what will happen to the average collision time, 1:? What will happen to the resistivity, p? Choose one: Hint: p = 2 where I is the collision time. nq r Explanation: When the temperature is increased, the
atoms in the medium are "vibrating" with faster average
speed. Free electrons will collide with atoms more
frequently. So the average collision time I is decreased.
The resistivity p=m/(nq2t), i.e. p is inversely proportional
to 1:. As the collision time decreases, resistivity increases.
Ans=2. 27.6.1 Power in two circuits case 2 V 3 identical bulbs are connected in two ways as
shown. Determine PII/PI, where PI is the power ofper bulb in case I, and P11 is in case 11.
Choose one: l
P11/P1 | 9 . 2
Hmt: PZWZIQRZV Explanation: ...
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- Fall '08