Suppose a 77.4 V battery is connected to a long cylindrical wire, and a current

Question

Suppose a 77.4 V battery is connected to a long cylindrical wire, and a current of 5.38 A flows. How many electrons will pass a given point in the wire every second? electrons What is the total resistance of the wire? ft If the wire is 54.1 m long, and its diameter is 0.0738 cm, find the resistivity of the wire. Find the work done by this battery to push one electron through the entire length of the wire. You will have three chances for this question. A cylindrical wire of length L has resistance R = 948 ft. The wire is now passed through a die so it is drawn out to length 3.11L.. What is the new resistance of the wire? HINT: The length is not the only quantity that has changed!

Explanation / Answer

a) 5.38*1/(1.6*10^-19) = 3.36*10^19

b) 77.4/5.38 = 14.4 ohm

c) resistibity = RA/L = 14.4*3.142*0.000738^2/(4*54.1) = 1.138*10^-7

d) W = eV = 1.609*10^-19*77.4 = 1.24*10^-17 J

e) LA = constant = L'A' = p

R = 948 = rL/A = rL^2/p

r = 948p/L^2

New resitance = R' = rL'/A' = rL'^2/p = (948/L^2)*3.11^2*L^2 = 948*3.11^2 = 9169.15 ohm

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

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