1. A bird stands on an electric transmission line carrying 2360A, as seen in the

Question

1. A bird stands on an electric transmission line carrying 2360A, as seen in the figure below.

The line has 2.60×10-5Ohm resistance per meter and the bird's feet are 4.40cm apart. What voltage does the bird feel? That is, what is the magnitude of the voltage difference between the bird's feet?

2.   A regular incandescent lightbulb (labeled 60W) might have a resistance of 14.7 Ohm when cold (20.0oC) and 171 Ohm when on (hot). Calculate its power consumption at the instant it is first turned on. (For this problem and the next, assume that this lightbulb experiences a fixed voltage difference of 110 V )

3. In the problem above, calculate the bulb's actual power consumption after a few moments when it is hot. This may be different than the label on the bulb indicates.

4.   Calculate the current i5 in the following configuration:

i1 = 9 A
i2 = 8 A
i3 = 4 A
i4 = 8 A

5. A number 16 copper wire has a diameter of 1.291 mm. If you measure the resistance to be 0.447 Ohm, how long must the wire be?

(Use rho=1.72x10-8Ohm-m for the resistivity of copper. )

6.   For safety, the National Electrical Code limits the allowable amount of current which such a wire may carry. When used in indoor wiring, the limit is 6 amperes for rubber insulated wire of that size. How much power would be dissipated in the wire of the above problem when carrying the maximum allowable current?

Explanation / Answer

1)

2.6*10^-5 ohms per meter x 0.04 meter = 1.04e-6 ohms

2360 A x 1.04e-6 ohms = 0.00245 volts.

2)

Assuming voltage is as rated on lighbulb:
Ic = 110/14.7 = 7.4 amps (inrush)
Ih = 110/171 = 0.64 amp (operating)

Power {when filament cold} = (Ic)²R = (7.4)²(14.7) = 804 watts ANS a)
Power {when filament hot} = (Ih)²R = (0.64)²(171) = 70 watts ANS b)

4)

according to Kirchhoff's Current Law the sum of currents entering some node in the circuit must be equal to the sum of the currents flowing away from that node (provided that the charge density in the circuit does not change in time) thus:

I1 + I2 + I4 = I3 + I5

hence, I5= I1 + I2 + I4 - I3

            =   9+8+8-4= 21 A

5)

Circumference of the circle, wire, 2(pi)r^2 and got that since the diameter was 1.291 mm it would be .001291 m and the radius would then be that divided by 2 so .0006455 m.

then doing 2(pi)r^2

1.6449E-5.

R=pl/A

Length of the wire= 43 m

6)

Power= i (current) X V
and Power= i^2 X R

so P=

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