Respuesta :
The equivalent resistance of n resistors connected in parallel is given by
[tex] \frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n} [/tex] (1)
In our problem, the resulting resistance of the 5 pieces connected in parallel is [tex]R_{Eq}=2.00 \Omega[/tex], and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
[tex] \frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R} [/tex]
From which we find [tex]R= 5 \cdot 2 \Omega = 10 \Omega[/tex].
So, each piece of wire has a resistance of [tex]10 \Omega[/tex]. Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
[tex]R_{Eq}=R_1 + R_2 + ...+R_n[/tex]
So if we apply it at our case, we have
[tex]R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega[/tex]
therefore, the resistance of the original wire was [tex]50 \Omega[/tex].
[tex] \frac{1}{R_{eq}} = \frac{1}{R_1}+ \frac{1}{R_2}+...+ \frac{1}{R_n} [/tex] (1)
In our problem, the resulting resistance of the 5 pieces connected in parallel is [tex]R_{Eq}=2.00 \Omega[/tex], and since the 5 pieces are identical, their resistance R is identical, so we can rewrite (1) as
[tex] \frac{1}{R_{Eq} }= \frac{1}{2 \Omega}= \frac{1}{R}+ \frac{1}{R} + \frac{1}{R} + \frac{1}{R} + \frac{1}{R} = \frac{5}{R} [/tex]
From which we find [tex]R= 5 \cdot 2 \Omega = 10 \Omega[/tex].
So, each piece of wire has a resistance of [tex]10 \Omega[/tex]. Before the wire was cut, the five pieces were connected as they were in series. The equivalent resistance of a series of n resistors is given by
[tex]R_{Eq}=R_1 + R_2 + ...+R_n[/tex]
So if we apply it at our case, we have
[tex]R_{eq}=R+R+R+R+R=5 R= 5\cdot 10 \Omega= 50 \Omega[/tex]
therefore, the resistance of the original wire was [tex]50 \Omega[/tex].
The resistance is indirectly proportional to the current flowing in the wire. The resistance of the given full-length wire will be 50Ω.
Parallel resistance can be given as:
[tex]\dfrac 1{Req} =\dfrac 1{R_1}+ \dfrac 1{R_2} + \dfrac 1{R_3} .....[/tex]
Where,
[tex]\dfrac 1{Req} [/tex] - resulting resistance = 2Ω
In the given question, the wire was cut down into 5 pieces.
So,
[tex]\dfrac 1 {2 \Omega } = \dfrac 5 {R}\\\\ R = 10 \Omega [/tex]
Resistance in the series of full length wire,
[tex]Req = R_1+ R_2+R_3 .........[/tex]
So,
[tex]R =R+R+R+R+R\\\\ R = 5 R\\\\ [/tex]
Put the value of [tex]R[/tex]
[tex]R eq = 5\times 10 \\\\ Req = 50 \Omega [/tex]
Therefore, the resistance of the given full-length wire will be 50Ω.
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