Question

On increasing the length of a wire, while keeping the area of cross section unchanged, the amount of current flowing through the wire shown by the ammeter

Answer 1

Answer:

The ammeter with show the decrease in current.

Explanation:

For instance,

Let the length of wire is doubled while keeping the area of cross-section unchanged.

We know,

$\text R = \rho \cfrac{\text l}{\text A}$

Let original resistance is $R_1$, length is $l_1$ and cross-section area is A.

$\therefore R_1 = \rho \cfrac{l_1}{A}$

Now, new length is double, i.e., $l_2 = 2l_1$ and new resistance is $R_2$

$\therefore R_2 = \rho \cfrac{l_2}{A}$

$= \rho \cfrac{2l_1}{A}$

$=2 \left( \rho \cfrac{l_1}{A} \right)$

$=2R_1$

$\text {Thus, } R_2 = 2 R_1$

Now, using Ohm's law, i.e., V = IR

Let original current flowing is $I_1$ for wire having resistance $R_1$ and having constant voltage V.

$\therefore I_1 = \cfrac{V}{R_1}$

Now, when length of wire is doubled while keeping the area of cross-section unchanged,

$I_2 = \cfrac{V}{R_2}$

Now, $R_2 = 2 R_1$

$\therefore I_2 = \cfrac{V}{2R_1}$

$\Rightarrow I_2 = \cfrac{1}{2} \left( \cfrac{V}{R_1} \right)$

$\Rightarrow I_2 = \cfrac{1}{2} I_1$

Thus, new current flowing will be half of originally flowing.

So, as the length of wire increases ammeter will show less current respectively.