A particular application requires a 47 m length of aluminum wire to have a 0.23

Question

A particular application requires a 47 m length of aluminum wire to have a 0.23 m ohm resistance at 20degreeC. What must be the wire's diameter? A platinum wire is connected to a battery. If the temperature increases, will the current in the wire increase, remain the same, or decrease? increases remains same decreases An electrical resistance thermometer is made of platinum wire that has a 4.9 ohm resistance at 20degreeC. The wire is connected to a 2.8 V battery. When the thermometer is heated to 2040degreeC, by how much does the current change?

Explanation / Answer

resistance =( resistivity*length)/area

resistivity=2.65*10^(-8)

length=47m

area=( resistivity*length)/resistance=5.42*10^(-3) m^2

area=pi(radius)^2=5.42*10^(-3)

radius=0.042 m

diameter=2*radius=0.083 m

(a) with the increase in temperature ,resistance increase and so current (= voltage/resistance) decreases

(b)initial current=2.8/4.9=0.5714A

R=R0(1+alpha(delta T)

R0=resistance at normal temperature=4.9

alpha=temperature coefficient =.003927

delta T=difference in temperature=2020

new resistance=R=43.77 ohm

new current=2.8/43.77=0.064 A

change by 0.5074 A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.