Question

a resistor is made out of a long wire having a length L. Each end of the wire is attached to a termina of a battery providing a constant voltage V0. A currect I flows through the wire. If the wire were cut in half, making two wires of length L/2, and both wires were attached to the battery (the end of both wires attached to one terminal, and the other ends attached to the other terminal), what would be the total currect flowing through the two wires?A) 4I B) 2I C) I D) I/2 E) I/4

Answer 1

Answer:

Option A

Solution:

As per the question:

Initially, voltage is $V_{o}$

Current is 'I' A

Length of the wire is L

Now,

We know that:

$R = \frac{\rho L }{A}$           (1)

where

$\rho = resistivity\ of\ the wire$

A = Cross sectional area of the wire

From eqn (1), if other things are taken to be constant, then

R ∝ L                 (2)

Thus

When the wire is cut into two reducing the length to $\frac{L}{2}$

Resistance, R' = $\frac{R}{2}$

Now, when these wires are connected as described, the connection is in parallel, therefore, the equivalent resistance of the two wires:

$\frac{1}{R_{eq}} = \frac{1}{R'} + \frac{1}{R'}$

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

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

Now, from Ohm's law:

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

Since, according to the question voltage $V_{o}$ is constant, thus

I ∝ $\frac{1}{R_{eq}}$

I ∝ $\frac{4}{R}$

Thus

I becomes 4I