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

Answer 1

Answer:

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 **twice** of the initial value when the length of wire was L.

Explanation:

V = IR

I = V/R

V = voltage or potential difference across the wire

I = current flowing in the wire

R = Resistance across the length of the wire

But the resistance is a function of the wire's physical attributes like its resistivity (ρ), cross sectional Area (A) and the length of wire (L)

R = ρL/A

If all other parameters are constant, R varies directly as the length of wire if all other parameters are constant

Let all the other parameters be k

R = kL

Initially, the length of the wire is L₁

R₁ = kL₁

I₁ = V/R₁

Then the wire is cut in half

L₂ = L₁/2

R₂ = kL₂

R₂ = kL₁/2 = R₁/2

I₁ = V/R₁

I₂ = V/R₂

Substituting R₂ = R₁/2 into the expression

I₂ = V/(R₁/2) = 2V/R₁ = 2 I₁

This means the current flowing in the wire of length L/2 is twice that flowing in the wire of length L.

Answer 2

The current will be double (twice) of that flowing in the original length of wire.

Given the following data:

• Current = 40 Amperes.

How to calculate the current.

Mathematically, current is given by this formula in accordance with Ohm's law:

$I = \frac{V}{R}$

Where:

• I is the current.

• R is the resistance.

• V is the voltage.

Mathematically, the resistance of a wire can be calculated by using this formula:

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

Note: The resistance (R) of a wire varies directly as its length (L) provided all other parameters and temperature are kept constant as follows:

$R=kL$

At the initial length, we have:

$R_1=kL_1\\\\I_1 = \frac{V}{R_1}$

When the length is halved, we have:

$L_2=\frac{L_1}{2}$

At this length, the resistance is given by:

$R_2=kL_2\\\\R_2=k\frac{L_1}{2} =\frac{R_1}{2}$

At the new length, the current is given by:

$I_2=\frac{V}{R_2} \\\\I_2=\frac{V}{\frac{R_1}{2} }\\\\\frac{2V}{R_1} =2I_2$

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