Question

an electric wire is stretched to increase its length by 25% by what percent will the resistance be increased and what will be the increase in its resistivity

Answer 1

Increase in length by 25%:

125%/100% × L = 1.25 L

Resistance is usually given by the formula :

The dimensions for the initial wire are:

R = kL/A

Length is L and a cross-sectional area of A

The new wire will have dimension of Length 1.25L and Cross-sectional area of A.

Calculate the resistance of the wire at first and after being stretched:

Initial resistance = kL/A
Final resistance = k1.25L/A

Find % increase = 1.25Lk/A ÷ kL/A × 100%
= 1.25kL/A × A/kL ×100%
=125%
125 % - 100% = 25%

There will be a 25 % increase in resistance.

Resistance is dependent on three factors:

1) Material from which material is made from. This is it's resistivity (k)and is a constant. It does not change in the same material

2) Length

3) Cross sectional area.

Answer 2

Answer:

25%

Explanation:

Increase in length by 25%:

125%/100% × L = 1.25 L

Resistance is usually given by the formula :

The dimensions for the initial wire are:

R = kL/A

Length is L and a cross-sectional area of A

The new wire will have dimension of Length 1.25L and Cross-sectional area of A.

Calculate the resistance of the wire at first and after being stretched:

Initial resistance = kL/A

Final resistance = k1.25L/A

Find % increase = 1.25Lk/A ÷ kL/A × 100%

= 1.25kL/A × A/kL ×100%

=125%

125 % - 100% = 25%

There will be a 25 % increase in resistance.

Resistance is dependent on three factors:

1) Material from which material is made from. This is it's resistivity (k)and is a constant. It does not change in the same material

2) Length

3) Cross sectional area.