
A metal wire of resistance \[R\] is cut into three equal pieces which are then connected side by side to form a new wire, the length of which is equal to one third of the original length. The resistance of this new wire is-
(1). \[R\]
(2). \[3R\]
(3). \[\dfrac{R}{9}\]
(4). \[\dfrac{R}{3}\]
Answer
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Hint: Resistance is directly proportional to the length and inversely proportional to area of cross section. As the wire is cut into three equal pieces, each will have a length equal to one-third of the original length. Determine the combination in which the pieces of wire are connected and calculate the resistance.
Formula used: \[r=\dfrac{\rho l}{A}\]
\[\dfrac{1}{R'}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}\]
Complete step by step answer:
The resistance of a material is the property by virtue of which it opposes the flow of current through it.
Resistance \[r\] of a wire of area of cross-section,\[A\] and length \[l\] and resistivity \[\rho \] is given by-
\[r=\dfrac{\rho l}{A}\] ---- (1)
As \[A\] and \[\rho \] are constant,
\[r\propto l\]
Let the resistance of each new piece be \[r'\], so,
\[\begin{align}
& \dfrac{r'}{R}=\dfrac{\dfrac{l}{3}}{l} \\
& \Rightarrow r'=\dfrac{R}{3} \\
\end{align}\]
Therefore the resistance of each new piece is \[\dfrac{R}{3}\] .
When the pieces are joined side by side they are now in a parallel combination, so the new wire formed by joining them will resistance equal to -
\[\dfrac{1}{R'}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}\]
\[\begin{align}
& \therefore \dfrac{1}{R'}=\dfrac{3}{R}+\dfrac{3}{R}+\dfrac{3}{R} \\
& \dfrac{1}{R'}=\dfrac{9}{R} \\
& R'=\dfrac{R}{9} \\
\end{align}\]
Therefore, the new wire formed by joining the pieces of the wire with resistance\[R\] has a resistance of \[\dfrac{R}{9}\].
So, the correct answer is “Option (3)”.
Note: Resistances can be connected in two ways- series combination or parallel combination. In series combination, the resistances are connected in a single path. Here, the resultant resistance is greater than the greatest resistance and the current is the same in all the resistances. In parallel combination, the resistances are connected in different branches between two same points. In parallel combination, the value of resultant is smaller than the smallest resistance and the voltage drop on each is the same.
Formula used: \[r=\dfrac{\rho l}{A}\]
\[\dfrac{1}{R'}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}\]
Complete step by step answer:
The resistance of a material is the property by virtue of which it opposes the flow of current through it.
Resistance \[r\] of a wire of area of cross-section,\[A\] and length \[l\] and resistivity \[\rho \] is given by-
\[r=\dfrac{\rho l}{A}\] ---- (1)
As \[A\] and \[\rho \] are constant,
\[r\propto l\]
Let the resistance of each new piece be \[r'\], so,
\[\begin{align}
& \dfrac{r'}{R}=\dfrac{\dfrac{l}{3}}{l} \\
& \Rightarrow r'=\dfrac{R}{3} \\
\end{align}\]
Therefore the resistance of each new piece is \[\dfrac{R}{3}\] .
When the pieces are joined side by side they are now in a parallel combination, so the new wire formed by joining them will resistance equal to -
\[\dfrac{1}{R'}=\dfrac{1}{{{R}_{1}}}+\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}\]
\[\begin{align}
& \therefore \dfrac{1}{R'}=\dfrac{3}{R}+\dfrac{3}{R}+\dfrac{3}{R} \\
& \dfrac{1}{R'}=\dfrac{9}{R} \\
& R'=\dfrac{R}{9} \\
\end{align}\]
Therefore, the new wire formed by joining the pieces of the wire with resistance\[R\] has a resistance of \[\dfrac{R}{9}\].
So, the correct answer is “Option (3)”.
Note: Resistances can be connected in two ways- series combination or parallel combination. In series combination, the resistances are connected in a single path. Here, the resultant resistance is greater than the greatest resistance and the current is the same in all the resistances. In parallel combination, the resistances are connected in different branches between two same points. In parallel combination, the value of resultant is smaller than the smallest resistance and the voltage drop on each is the same.
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