Question

A wire of 15 ohm resistance is gradually stretched to double it's original length.it is then cut into two equal parts. These parts are then connected in a parallel across a 3 volt battery.find the current drawn from the battery.

Answer 1

Hi .


Let R0, l0, A0 be the original resistance, length and area of the wire.

Let R, l, A be the original resistance, length and area of the stretched wire. Since the length is doubled ; l =2l0

When the wire is stretched its Volume remains the same.

Therefore, V= Al = A0l0

A(2l0) = A0l0

A=A0/2

Resistance R=ρl/A

R/R0 = lA0/l0A

R/R0 = 2l0 A0/ (l0 (A0/2)) = 4

R = 4R0 = 4x15 = 60Ω

When the resistance of wire R=60Ω is cut into two equal parts, the length becomes half but area remains the same. Hence resistance of each part is R­1= R/2 = 60/2 = 30Ω..

When these two wires of resistance 30Ω are connected in parallel across a 3V battery, the equivalent resistance Rp = R1/2 = 30/2 =15 Ω

(1/Rp = (1/ 30) +( 1/ 30) = 2/30 = 1/15

Rp =15Ω )

Therefore the current through the circuit is V/Rp = 3/15 = 0.2A



Hope this helps u!

Answer 2

Concept introduction:

Resistance:  A force, such as friction, that acts in the opposing direction of a body's motion and tends to stop or slow it down. The amount that a material obstructs the passage of electric current caused by a voltage.

Given:

The resistance of the wire,  $R=15$Ω

The original length of the conductor,  $l_{1}$

The final length of the conductor,   $l_{2} =3l_{1}$

The voltage of the battery, $V=3V$

To find:

We have to find the current drawn from the battery.

Solution:

According to the problem,

$R=\frac{pl}{A}$

Putting the values we get $R^{'} =4R$

                                          $=R*4\\=4*15=60$Ω

∴ $\frac{1}{R} =\frac{1}{R_{1} } +\frac{1}{R_{2} } \\- > 15$

The current drawn from the battery is

$I=\frac{V}{R}\\=\frac{3}{15}\\ 0.2$

Final answer:

So. the the current drawn from the battery is $0.2A$.

#SPJ2