Question

A bird stands on a dc electric transmission line carrying 2900 A. The line has 2.6×10^-5 resistance per meter, and the bird's feet are 3.9 cm apart. What is the potential difference between the bird's feet?
Express your answer to two significant figures and include the appropriate units

Answer 1

Concepts and reason

The concept is required to find the potential difference between the bird’s feet is Ohm’s law.

First find the resistance of the electric transmission line by using the length of the electric transmission line and resistivity of the electric transmission line to find the potential difference between the bird’s feet.

Fundamentals

The resistance of the resistance of the electric transmission line is,

$R = \rho l$

Here, $\rho$is the resistivity of the electric transmission line and l is the length of the electric transmission line.

Use Ohm’s law to find the potential difference between the bird’s feet.

$V = IR$

Here, I is the current carrying by dc electric transmission line and R is the resistance of electric transmission line.

The resistance of the resistance of the electric transmission line is,

$R = \rho l$

Here, $\rho$is the resistivity of the electric transmission line and l is the length of the electric transmission line.

Substitute 0.039m for l and $2.6 \times {10^{ - 5}}\Omega /m$ for $\rho$in the above equation

$\begin{array}{c}\\R = \left( {2.6 \times {{10}^{ - 5}}\Omega /m} \right)\left( {0.039m} \right)\\\\ = 1.014 \times {10^{ - 6}}\Omega \\\end{array}$

Hence, the resistance of the electric transmission line is$1.014 \times {10^{ - 6}}\Omega$.

Use Ohm’s law to find the potential difference between the bird’s feet.

$V = IR$

Here, I is the current carrying by dc electric transmission line and R is the resistance of the electric transmission line.

Substitute $1.014 \times {10^{ - 6}}\Omega$for R and 2900A for I in the above equation.

$\begin{array}{c}\\V = \left( {1.014 \times {{10}^{ - 6}}\Omega } \right)\left( {2900A} \right)\\\\ = 2.9 \times {10^{ - 3}}V\\\\ = 2.9mV\\\end{array}$

Therefore, the potential difference between the bird’s feet is$2.9mV$

Ans:

The potential difference between the bird’s feet is$2.9mV$