Question

A typical overhead power line is bare wire at a high potential relative to ground. A local line receives power at

3300 V with respect to the ground; the delivered voltage is slightly less, due to resistance loss in the wire. The

wire is 10 km long, carries a current of 60 A, and has a resistance (made small to limit power loss) of 2.5 Ω.

1) Approximately how much power is lost due to the resistance of the wire?

A. 9 kW

B. 50 kW

C. 200 kW

D. 4.5 MW

2) If the diameter of the wire is doubled, how does the power lost due to the resistance of the wire change?

(Assume that the current and the voltage stay the same.)

A. It increases by a factor of 2.

B. It stays the same.

C. It decreases by a factor of 2.

D. It decreases by a factor of 4.

A bird can safely perch on the wire because the voltage drop along the wire leads to a very small potential

difference between the bird’s feet. (Touching a wire while standing on the ground would lead to a different result!)

A large bird is perched on the wire, with feet 6.0 cm apart.

-What is the potential difference between the two ends of the wire?

- What is the electric field in the wire corresponding to this potential difference?

- What is the difference in potential between the bird’s feet?

Answer 1

1.

Power lost due to resistance of the wire is

P=I²R =60²*2.5

P=9 KW

2.

Resistance is given by

R=pL/A =4pL/pid²

if diameter is doubled ,Resistance of wire decreases by a factor of 4 ,therefore power lost decreases by a factor by 4.

Answer is D. It decreases by a factor of 4.

3.

a)

Potential difference between the ends of the wire is

V=IR =60*2.5=150 Volts

b)

E=V/d =150/10000=0.015V/m

c)

R₂/R₁=L₂/L₁

R₂=2.5*(0.06/10000)=1.5*10^-5ohms

V₂=60*1.5*10^-5=9*10^-4V=0.9 mV