Question

A wire of given material having length ‘L’ and area of cross section ‘A’ has a resistance of 16 Ohms. What would be the resistance of another wire of the same material having length ‘L/2’ and cross-sectional area ‘2A’.​

Answer 1

EXPLANATION.

• GIVEN

A wire of natural material having length = L

area of cross section = A

Resistance = 16 ohms

if the resistance of another wire of the same

material having length = L/2

Area of cross section = 2A

To find the new resistance

According to the question,

Formula of resistance

$\bold{\frac{ \rho \: l}{a} = resistance \: \: = \rho \: \: is \: \: constant}$

For first wire,

Length = L

Cross section area = A

Resistance = 4 ohms

$\bold{\frac{ \rho \: l}{a} = 4}$

For second wire,

Length = L/2

area of cross section = 2A

$\bold{resistance \: = \frac{ \rho \: l}{4a} }$

$\bold{given \: = \frac{ \rho \: l}{a} = 4 }$

put the value of resistance in second wire,

Therefore,

4 / 4 = 1 ohms

Therefore,

Resistance of second wire = 1 ohms

Answer 2

Explanation:

EXPLANATION.

GIVEN

A wire of natural material having length = L

area of cross section = A

Resistance = 16 ohms

if the resistance of another wire of the same

material having length = L/2

Area of cross section = 2A

To find the new resistance

According to the question,

Formula of resistance

\bold{\frac{ \rho \: l}{a} = resistance \: \: = \rho \: \: is \: \: constant}

a

ρl

=resistance=ρisconstant

For first wire,

Length = L

Cross section area = A

Resistance = 4 ohms

\bold{\frac{ \rho \: l}{a} = 4}

a

ρl

=4

For second wire,

Length = L/2

area of cross section = 2A

\bold{resistance \: = \frac{ \rho \: l}{4a} }resistance=

4a

ρl

\bold{given \: = \frac{ \rho \: l}{a} = 4 }given=

a

ρl

=4

put the value of resistance in second wire,

Therefore,

4 / 4 = 1 ohms

Therefore,

Resistance of second wire = 1 ohms