Question

 

1. The resistance of a very fine aluminum wire with a 10 μm × 10 μm square cross section is 1200 Ω . A 1200 Ω resistor is made by wrapping this wire in a spiral around a 3.4-mm-diameter glass core.

Part A: How many turns of wire are needed? Express your answer using two significant figures.

2.The starter motor of a car engine draws a current of 180 A from the battery. The copper wire to the motor is 5.40 mm in diameter and 1.2 m long. The starter motor runs for 0.680 s until the car engine starts.

Part A: How much charge passes through the starter motor? Express your answer with the appropriate units.

Part B: How far does an electron travel along the wire while the starter motor is on? Express your answer with the appropriate units.

3. A capacitor is discharged through a 100 Ω resistor. The discharge current decreases to 29.0% of its initial value in 3.00 ms . What is the value of the capacitor? Express your answer with the appropriate units.

Answer 1

1)

R = 1200 Ω

ρ = 2.65*10^-8 Ωm (Resistivity of aluminum - a constant)

A = (10 μm)^2 = (10*10^-6 m)^2 = 1*10^-10 m^2

L = (R*A) / ρ

L = (1200 Ω * 1*10^-10 m^2) / (2.65*10^-8 Ωm)

L = 4.528m

C = Circumference of wire
= 2π*r = 2π*(1.7*10^-3 m) = 0.01068 m

Turns = L / C

Turns = (4.528 m) / (0.01068 m)

Turns = 423.9

2significant figure = 420

2)

(a)

The current, multiplied by the duration gives you the number of coulombs of charge that have passed:

Q = I*t = 180*0.680 = 122.4 C

(b)

The radius of the conductor is 2.7mm, or .0027m

The area of the conductor is:

A = π*r^2 = π*(.0027)^2 = 2.29*10^-5 m^2

The current density is:

J = 180/2.29*10^-5 = 7860262.00A/m

According to the listed reference:

Vd = J/(n*e) = 7860262.00 / ( 8.46*10^28 * 1.6*10^-19 ) =0.00058
= 0.580 mm/s

The distance traveled is:

x = v*t = 0.580 * .680 = 0.522 mm

3)

The discharge current decreases to 29% of its initial value because the charge on the capacitor (and its voltage) have decreased to 29% of their initial values. You can solve using the equation

Q = Qo(e)^(-t/RC)

Where,
Q= charge on the capacitor at time t
Qo = initial charge
R = resistance in ohms
C = capacitance in Farads
t = 0.003 second

Q = Qo(e)^(-0.003/100C)

We know that Q= 0.29Qo, so we can make that subsitution and eliminate the Qo's. All that's left is to solve for C.

0.29Qo = Qo(e)^(-0.003/100C)
0.29 = (e)^(-0.003/100C), then take the natural log of both sides
ln(0.29) = -0.003/100C
100C = -0.003/ln(0.29)
100C = 0.0024
C = 0.0000242 Farad, or 24.2uF