Question

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? 2I 4I I/2 I/4 I

Answer 1

Answer: 4I

Explanation:

Ohms law states that the current in a given conductor is directly proportional to the voltage and inversely proportional to the resistance in the wire.

Resistivity is that property of a material that determines how strong it resists electric current.

first we make the resistance of the wire of length L be R Ω.

Using ohms law V = IR

Then, I = V/R, I = V•/R

Secondly, we recall that the resistivity of a wire is found using the formula R = ρL/A

The new resistance of each length i.e. L/2 will be equal to R/2 Ω, and the resistors are now connected in parallel:

Using

1/R(total) = 1/R1 + 1/R2

1/R(total) = 2/R + 2/R

R(total) = R/4

Going back to use ohms law,

V• = i2 * R/4

i2 = 4*V•/R

Recall that I = V•/R

i2 = 4I

Answer 2

Answer:

Option B. The current will be 4I

Explanation:

When the ends of the wire are attached across the terminals of a battery

Resistance = R

Voltage = V₀

According to Ohm's law, V = IR

I₁ = V₀ /R --------------(i)

When the wire was cut into halves

Since the length has been halved, the resistance will also be halved because length ∝ Resistance

The resistance of each length becomes R/2

The description of the connection indicates that the resistors are connected in parallel.

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

$R_{eq} = \frac{R}{4}$

Applying Ohm's law, V = IR

$V_{o} = I_{2} (\frac{R}{4} )\\I_{2} = (\frac{4}{R} )V_{o}$

$I_{2} = (\frac{V_{o} }{R})4\\But I_{1} = \frac{V_{o} }{R}\\I_{2} = 4 I_{1}$