
How will the resistance of wire change when: Its length is doubled?
Answer
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Hint:We will be using the expression of resistivity of wire which is a constant value and expressed in the form of resistance of the wire, cross-section area of the wire and length of the wire. We will establish the relationship between final and initial resistance so that a conclusion can be made.
Complete step by step answer:
Resistivity is the property of a conductive material which resists the flow of current when it is subjected to a voltage.
\[\rho = \dfrac{{RA}}{l}\]
Here, R is the resistance of the given wire, l is the length of the wire and A is area of cross-section of the wire.
Rewriting the above expression for resistance of the wire.
\[R = \dfrac{{\rho l}}{A}\]
Let us write the new equation of resistance when length is doubled.
\[R' = \dfrac{{\rho l'}}{{A'}}\]……(2)
We know that the volume of the wire is equal to the ratio of the product of cross-sectional area and length of the wire.
\[V = A \cdot l\]
When the length of wire changes keeping volume constant, area of the wire will also change and that change can be calculated as below.
\[V = A\left( {2l} \right)\]
But we know that the volume of wire is to be kept constant so to accommodate the increase in length cross-area is to be reduced to half of the actual cross sectional area.
Substitute \[\dfrac{A}{2}\] for \[A'\] and \[2l\] for \[l'\] in equation (2) to get the value of new resistance.
\[\begin{array}{c}
R' = \dfrac{{\rho \left( {2l} \right)}}{{\dfrac{A}{2}}}\\
= 4\left( {\dfrac{{\rho l}}{A}} \right)
\end{array}\]
Substitute R for \[\left( {\dfrac{{\rho l}}{A}} \right)\] in the above expression.
\[R' = 4R\]
Therefore, we can say that by doubling the length of the given wire its resistance will become four times of the initial value of resistance.
Note:We have to remember that the volume of the wire is constant then only further calculations could be done. Also it has to be noted that the resistivity is the property of the material that means it does not depend on the area and length of the wire hence it is kept constant.
Complete step by step answer:
Resistivity is the property of a conductive material which resists the flow of current when it is subjected to a voltage.
\[\rho = \dfrac{{RA}}{l}\]
Here, R is the resistance of the given wire, l is the length of the wire and A is area of cross-section of the wire.
Rewriting the above expression for resistance of the wire.
\[R = \dfrac{{\rho l}}{A}\]
Let us write the new equation of resistance when length is doubled.
\[R' = \dfrac{{\rho l'}}{{A'}}\]……(2)
We know that the volume of the wire is equal to the ratio of the product of cross-sectional area and length of the wire.
\[V = A \cdot l\]
When the length of wire changes keeping volume constant, area of the wire will also change and that change can be calculated as below.
\[V = A\left( {2l} \right)\]
But we know that the volume of wire is to be kept constant so to accommodate the increase in length cross-area is to be reduced to half of the actual cross sectional area.
Substitute \[\dfrac{A}{2}\] for \[A'\] and \[2l\] for \[l'\] in equation (2) to get the value of new resistance.
\[\begin{array}{c}
R' = \dfrac{{\rho \left( {2l} \right)}}{{\dfrac{A}{2}}}\\
= 4\left( {\dfrac{{\rho l}}{A}} \right)
\end{array}\]
Substitute R for \[\left( {\dfrac{{\rho l}}{A}} \right)\] in the above expression.
\[R' = 4R\]
Therefore, we can say that by doubling the length of the given wire its resistance will become four times of the initial value of resistance.
Note:We have to remember that the volume of the wire is constant then only further calculations could be done. Also it has to be noted that the resistivity is the property of the material that means it does not depend on the area and length of the wire hence it is kept constant.
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