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# Electric Resistance and Ohm's Law

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A bird lands on a bare copper wire carrying a current of 32 A. The wire is 8 gauge, which means that its cross-sectional area is 0.13 cm^2. (a) Find the difference in potential between the bird's feet, assuming they are separated by a distance of 6.0 cm. (b) Will your answer to part (a) increase or decrease if the separation between the bird's feet increases? Explain

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A bird lands on a bare copper wire carrying a current of 32 A. The wire is 8 gauge, which means that its cross-sectional area is 0.13 cm^2. (a) Find the difference in potential between the bird's feet, assuming they are separated by a distance of 6.0 cm. (b) Will your answer to part (a) increase or decrease if the separation between the bird's feet increases? Explain

(a) Find the difference in potential between the bird's feet, assuming they are separated by a distance of 6.0 cm.

Step 1: Find the resistance of the wire between the bird's feet

R = rho L / A
where
rho= resistivity
l= length of wire
A= cross sectional area of wire

In this case
L= 6 cm= 0.06 m
A= 0.13 cm^2= 0.000013 m^2=
the resistivity of copper wire rho= 0.0000000172 ohm -m

therefore resistance= 0.000079 ohm =0.0000000172*0.06/0.000013

Step 2: Find the potential difference

V= IR
where
V= potential difference
I=current
R= resistance

In this case
I= 32 A
R= 0.000079 ohm

Therefore V= IR= 0.00253 V =32*0.000079

Potential difference= 0.00253 V

(b) Will your answer to part (a) increase or decrease if the separation between the bird's feet increases? Explain

The answer to part (a) will increase
This is because resistance is directly proportional to the length of wire.
Thus if the separation (that is length) increases, the resistance of the wire between the feet would also increase

Since the potential difference = I R ,and the current(I) does not change while the resistance [R] increases , the potential difference would also increase

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