Now let us take a look at the cables used to deliver power from the battery to s
ID: 2091864 • Letter: N
Question
Now let us take a look at the cables used to deliver power from the battery to starter in the previous problem. We will need two copper wires, each 1.0 m long. (An automotive starter motor is designed to draw a current of 100 A from an ideal voltage source of 12V. The starter is powered by a real car battery with an internal resistance of 0.01 Ohms.) this the previous question What is the total resistance of the cables if we choose to use wires with a diameter of 1.29 mm (American Wire Gauge #16)? What is the current drawn by the starter if we use the cables described above? What is the electrical power dissipated in the cables when the starter is engaged? How much power is delivered to the starter? What is the minimum diameter of wires we need to always be able to start this car (1000 Watts delivered to starter)?Explanation / Answer
Here ishowtosolve ALL parts of the problem in detail...
a. The resistance of any wire is given by R = P * l/A where p is the constant of proportionality, l is the length of the wire and A is the cross-sectional area...
For copper, P = 1.68*10^-8 (According to Giancolli Volume 2) the cross sectional area is r^2 pi
So R = (1.68*10^-8)(1.0)/(.645^2*pi) = .013 ohms,
since their are 2 wires the total resistance of the wires is .026 Ohms
b. In order to find the current drawn by the starter we must use the equation I = V/(R+r+Rwires) = 12/(.12+.01+.026) = 76.9 Amps
** R is the resistance of the starter which you should have found in the previous problem to be .12 and r is the internal resistance of the battery, stated in the problem
c. The electrical power dissipated can be found using P=I^2Rwires ... P = (76.9)^2(.026) = 153.75 watts
D. In order to find the power delivered to the starter we must find the terminal voltage by subtracting the current * the internal resistance from the voltage of the battery...
Vt = V - Ir = 12 - (76.9)(.01) = 11.231 volts
Multiplyingthe terminal voltage by the currentwill give usthe power delivered to the starter...
P = Vt *I = (11.2318*76.9) = 863.66 Watts
E. As we can see from part d, the given diameter is not enough to start the car... First we need to find the minimum current to start the car...
P = I^2 R
1000 = I^2 * .12 so the minimum current to start the car is ...I = 91.28 A
Now using I = V/(R+r+Rwires) we can solve for Rwires...
91.28 = 12 /(.12+.01 + Rwires)
Rwires = .00145 ohms
Since that is the resistance of both wires.. the resistance in one wire is .00145/2 = .000726 Ohms
Using the equation from above R = (1.68*10^-8)(1.0)/(radius^2*pi) where R= .000726 we can solve for radius.
[(1.68*10^-8)(1.0)/(.000726)] * pi = radius^2,
radius = .0027meters Diameter = 2*radius = 2*.0027 = .0054 meters or 5.4mm
Thus the minimum diameter of the wires should be at least 5.4 mm in order to start the car.
So itmakes sense that in Part D the wires delivereda power less than 1000 watts and were not able to start the carbecause the diameter of the wires was not big enough.
Hope that helped :)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.