Question

A 3 ohm resistor stretched to twice its original length find new resistance

Answer 1

Given:

A 3 ohm resistor stretched to twice its original length

To Find:

New resistance

Solution:

We know that,

• Resistance R of a resistor is given by

$\orange{\boxed{\bf{R=\dfrac{\rho l}{A}}}}$

where,

ρ is the resistivity

l is length

A is Area

$\rule{190}{1}$

Let the initial resistance of resistor be R, initial length be l, ρ be the resistivity and A be the area

So,

$\longrightarrow\rm{R=\dfrac{\rho l}{A}}------(1)$

$\rule{190}{1}$

Now, after stretching the resistor to twice its original length, new resistance will be R' and new length will be 2l

So,

$\longrightarrow\rm{R'=\dfrac{\rho (2l)}{A}}$

$\longrightarrow\rm{R'=\dfrac{2\rho l}{A}}$

$\longrightarrow\rm{R'=2\bigg(\dfrac{\rho l}{A}\bigg)}$

From (1), we get

$\longrightarrow\rm{R'=2(R)}$

$\longrightarrow\rm\green{R'=2R}$

Hence, the new resistance is twice the initial resistance of resistor.

Answer 2

$\bigstar$ Answer:

Given ,

Resistance , R1 = 3 Ω

Resistance , R2 = ?

Length , l1 = 2l

Length , l2 = l

resistivity and are remains constant here.

$\bigstar$ Formula Used :

• $\bold{R=\frac{\rho l}{A} }$

• $\bold{\rho=\frac{RA }{l} }$

$\bigstar$ Explanation :

Now ,

$\rho_1=\rho_2\\\\\implies \frac{R_1l_1}{A_1}=\frac{R_2l_2}{A_2}\\\\ \implies \frac{3*2l}{A}=\frac{R_2*l}{A}\\\\ \implies \bold{\bf{\red{R_2=6\Omega}}}\\\\\implies R_2=2(R_1)$

Hence the new resistance is twice the Original one