Question

The resistance of a wire is 15 ohm, the wire is drawn out, so that it's length becomes 3 times the original length. Calculate the new resistance in this situation​

Answer 1

AnsWer :

135Ω

To FinD :

Calculate the new resistance in this situation

SolutioN :

We have,

• 15Ω resistance.
• Wire cut so that it's length becomes 3 times the original length.

Now,

• We can also write it's relation.
• When drawn a wire 3 times the original then, length of wire be 3L and Area become A / 3.

Let,

• Original Area of wire be A
• change after wire drawn out be A'
• Original Length of wire be L
• Change after wire drawn out be L'

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)   \put(1,2){\line(2,0){15}}\put(1,4){\line(2,0){15}}\put(4.5,-2){ \tt{L\:,\: A} }\qbezier(1,2)(0,3)(1,4)\qbezier(16,2)(17,3)(16,4)\end{picture}$

We know,

$\tt \dagger \: \: \: \: \: \boxed{\tt{R = \rho \dfrac{L}{A}} }$

✐ For Original Wire :

$\tt \dagger \: \: \: \: \: R = \rho \dfrac{L}{A} \: \: \: \: \: - (1)$

✐ After Wire is drawn out.

$\tt \dagger \: \: \: \: \: R' = \rho \dfrac{L'}{A'} \: \: \: \: \: - (2)$

☛ By Equation ( 1 ) and ( 2 )

$\tt \dagger \: \: \: \: \: \dfrac{ R'}{R}= \frac{ \rho \dfrac{L'}{A'}}{\rho \dfrac{L}{A}}$

$\tt \dagger \: \: \: \: \: \dfrac{ R'}{R}= \dfrac{L'}{A'} \times \dfrac{A}{L}$

$\tt \tt : \implies \dfrac{ R'}{R}= \dfrac{A}{A'} \times \dfrac{L'}{L}$

$\tt \tt : \implies \dfrac{ R'}{R}= \dfrac{3}{1} \times \dfrac{3}{1}$

$\tt : \implies\dfrac{ R'}{R}= 9$

$\tt : \implies\dfrac{ R'}{15}= 9$

$\tt : \implies R'= 9 \times 15.$

$\tt : \implies R'= 135\Omega.$

✬ Therefore, the value of new resistance be 135Ω.

$\rule{200}3$

MorE InformatioN :

R = ρ L / A.

Where as,

• ( R ) Resistance.
• ( ρ ) Resistivity
• ( L ) Length.
• ( A ) Area.

Answer 2

AnsWer :

135Ω

To FinD :

Calculate the new resistance in this situation

SolutioN :

We have,

15Ω resistance.

Wire cut so that it's length becomes 3 times the original length.

Now,

We can also write it's relation.

When drawn a wire 3 times the original then, length of wire be 3L and Area become A / 3.

Let,

Original Area of wire be A

change after wire drawn out be A'

Original Length of wire be L

Change after wire drawn out be L'

\setlength{\unitlength}{1mm}\begin{picture}(5,5) \put(1,2){\line(2,0){15}}\put(1,4){\line(2,0){15}}\put(4.5,-2){$\tt{L\:,\: A}$}\qbezier(1,2)(0,3)(1,4)\qbezier(16,2)(17,3)(16,4)\end{picture}

We know,

$\tt \dagger \: \: \: \: \: \boxed {\tt \pink{R = \rho \dfrac{L}{A}} }† </p><p>$

✐ For Original Wire :

$\tt \dagger \: \: \: \: \: R = \rho \dfrac{L}{A} \: \: \: \: \: - (1)†R=ρ </p><p>$

✐ After Wire is drawn out.

$\tt \dagger \: \: \: \: \: R' = \rho \dfrac{L'}{A'} \: \: \: \: \: - (2)</p><p>$

☛ By Equation ( 1 ) and ( 2 )

\tt \dagger \: \: \: \: \: R' = \rho \dfrac{L'}{A'} \: \: \: \: \: - (2)

$</p><p>\tt \dagger \: \: \: \: \: \dfrac{ R'}{R}= \dfrac{L'}{A'} \times \dfrac{A}{L}† </p><p>$

$</p><p>\tt \tt : \implies \dfrac{ R'}{R}= \dfrac{A}{A'} \times \dfrac{L'}{L}</p><p>$

$\tt \tt : \implies \dfrac{ R'}{R}= \dfrac{3}{1} \times \dfrac{3}{1}$

$\tt : \implies\dfrac{ R'}{R}= 9$

$\tt : \implies\dfrac{ R'}{15}= 9$

$\tt : \implies R'= 9 \times 15.$

$\tt : \implies R'= 135\Omega.$

✬ Therefore, the value of new resistance be 135Ω.

MorE InformatioN :

R = ρ L / A.

Where as,

( R ) Resistance.

( ρ ) Resistivity

( L ) Length.

( A ) Area.