Question

When a wire that has a large diameter and a length L is connected across the terminals of an automobile battery, the current is 40 A. If we cut the wire to half of its original length and connect one piece that has a length L/2 across the terminals of the same battery, the current will be:

A) 160A
B) 40A
C) 10A
D) 80A
E) 20A

Thank you so much!

Answer 1

Concepts and reason

The concept required to solve this problem is Ohm’s law and resistivity.

Initially write the relation of resistance, length, resistivity, and area. Then substitute this value of resistance in the equation of Ohm’s law.

Fundamentals

The resistivity of the material is a measure of its resisting power. It is expressed as,

$\rho = \frac{{RA}}{L}$

Here, $\rho$ is the resistivity, R is the resistance, A is the area of cross-section, and L is the length.

The Ohm’s law states that the voltage in a battery is directly proportional to the current flowing in the circuit and the resistance offered by the resistor. It is given by,

$V = IR$

Here, V is voltage, I is the current, and R is the resistance.

The expression of resistivity is given by,

$\rho = \frac{{RA}}{L}$

Rearrange the equation to find the value of resistance.

$R = \frac{{L\rho }}{A}$

According to the given condition, the length became half. So, the new resistance $R'$ will be,

$R' = \frac{{\left( {\frac{L}{2}} \right)\rho }}{A}$

By symmetry, the resistance will also become half.

$R' = \left( {\frac{1}{2}} \right)\left( {\frac{{\rho L}}{A}} \right)$

The Ohm’s law is expressed as,

$\begin{array}{l}\\V = IR\\\\I = \frac{V}{R}\\\end{array}$

So, current I is inversely proportional to R.

Thus,

$I \propto \frac{1}{R}$

So, if R is halved, I will be twice.

$2I \propto \frac{2}{R}$

Here, 40 A is I. Thus, the new current will be 2I.

So, the current is 80 A.

Ans:

The current is 80 A.