Question

6. A resistor is made out of a wire having a length L. When the ends of the wire are attached across the terminals of an ideal battery having a constant voltage V0 across its terminals, a current I flows through the wire. If the wire were cut in half, making two wires of length L/2, and both wires were attached across the terminals of the battery (the right ends of both wires attached to one terminal, and the left ends attached to the other terminal), how much current would the battery put out?

Answer 1

Answer:

4 I

Explanation:

The initial resistance of the wire is given by:

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

where

$\rho$ is the resistivity of the material

L is the length of the wire

A is the cross-sectional area of the wire

Since the resistance is proportional to the length of the wire, when the wire is cut in half, each wire will have half of the initial resistance:

$R' = \frac{\rho \frac{L}{2}}{A}=\frac{R}{2}$

Later, these two pieces of wire are connected in parallel to the battery (because the right ends of both wires are attached to one terminal, while the left ends are attached to the other terminal, so they are connected in parallel). Therefore, their total resistance will be:

$\frac{1}{R_T} = \frac{1}{R_1}+\frac{1}{R_2}=\frac{1}{R/2}+\frac{1}{R/2}=\frac{2}{R}+\frac{2}{R}=\frac{4}{R}\\R_T = \frac{R}{4}$

so the total resistance of the circuit will be 1/4 of the initial resistance.

The initial current in the circuit was given by Ohm's law:

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

Since the voltage V has not changed, the new resistance will be

$I' = \frac{V}{R_T}=\frac{V}{R/4}=4 \frac{V}{R}=4I$

so, the current will increase by a factor 4.

Answer 2

When the length of the wire is halved and connected in parallel, then the current increases four times the initial current.

What is Ohm's law?

Ohm's law states that when there is no physical change in the conductor, the potential difference is directly proportional to the current flowing in it.

V = RI

V be the potential difference

I be the current.

R be the resistance of the circuit.

A resistor is made out of a wire having a length of L.

When the ends of the wire are attached across the terminals of an ideal battery having a constant voltage V₀ across its terminals, a current I flows through the wire.

We know that the resistance is given by

$\rm R = \dfrac{\rho L}{A}$

where L is the length, A is the area, and ρ is the resistivity of the conductor.

If the wire were cut in half, making two wires of length L/2, and both wires were attached across the terminals of the battery. Then we have

$\rm R '= \dfrac{\rho \frac{L}{2}}{A} = \dfrac{R}{2}$

These two wires are in parallel. Then we have

$\rm \dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_1}\\\\\\\dfrac{1}{R_{eq}} = \dfrac{1}{\frac{R}{2}}+\dfrac{1}{\frac{R}{2}}\\\\\\R_{eq } \ = \dfrac{R}{4}$

Then the current will be given by Ohm's law

$\rm I' = \dfrac{V}{R_{eq}}\\\\\\I' = \dfrac{V}{\frac{R}{4}} \\\\\\I' = 4 \dfrac{V}{R} \\\\\\ I ' = 4 \ I$

More about Ohm's law link is given below.

https://brainly.com/question/796939