Question

In how many parts(equal) a wire of 100 Ω be cut so that a resistance of 1 Ω is obtained by connecting them in parallel ?

a) 10

b) 5

c) 100

d) 50

Answer 1

@Fifciol

Answer 2

@UnkleRhaukus

Answer 3

What resistance what do you get when you cut the wire in half?

Answer 4

two resistors will be \(R=50\Omega\)

in parallel \(R_T=\left(\frac1{50\Omega}+\frac1{50\Omega}\right)^{-1}\\=(2/50\Omega)^{-1}\\=25\Omega\)

Answer 5

then

Answer 6

four resistors wil be \(25\Omega\)\[R_T=\left(\frac1{25\Omega}+\frac1{25\Omega}{+\frac1{25\Omega}+\frac1{25\Omega}+}\right)^{-1}\\=(4/25\Omega)^{-1}\\=6.25\Omega\]

Answer 7

@UnkleRhaukus its a long process

Answer 8

there are only four options

Answer 9

and if there are no options, what can we do?

Answer 10

@Festinger

Answer 11

the resistance is proportional to length
\[R=\rho \frac{ l }{ A }\]
the resistance of one wire of length l- R1=100
If you cut it in a half:
\[\frac{ R_1 }{ l }=\frac{ R_2 }{ \frac{l }{ 2 } } \rightarrow R_2=\frac{ R_1 }{ 2 }\]
if you cut it in n parts the resistance will be
\[R=\frac{ R_1 }{ n }\]
because you shorten the length n times so it became \[\frac{ l }{ n }\]
When you connect these parts in parallel the output resistance must be 1 ohm so
\[1=\frac{ 1 }{ R }+\frac{1 }{ R }+...=n\frac{ 1 }{ R }=n\frac{1 }{ \frac{ R_1 }{ n } }=\frac{n^2 }{ R_1 }=\frac{ n^2 }{ 100 } \rightarrow n=10\]

Answer 12

something unclear?

Answer 13

i don't understand

Answer 14

can you explain me? @Fifciol from starting point

Answer 15

in first equation A is the area of wire which remains constant even if you cut it. rho is restivity which also remains constant when you cut it because you don't change the material that wire is made of. The ratio
\[\frac{ R}{ l }=\frac{ \rho }{A }\]
must be therefore always the same

Answer 16

so the resistance of wire that hasn't been cut yet is \[\frac{ 100 ohms }{ l }\]
I arbitrally chose l as length. the resistance of wire per unit length, which is cut into n parts is\[\frac{R }{ \frac{ l }{ n } }=\frac{ nR }{ l }\]
so
\[\frac{ 100}{ l }=\frac{ nR }{ l } \rightarrow R=\frac{100 }{ n }\]

Answer 17

your output resistance must be 1 ohm, and they have to be connected in parallel so
\[\frac{ 1 }{ 1ohm }=\frac{ 1 }{ R }+\frac{ 1 }{R }+...\]
where R is that resistance of cut wire

Answer 18

then

Answer 19

we can see if there were 3 parts your inverse resistance would be
\[\frac{ 1 }{ R }+\frac{ 1 }{ R }+\frac{1 }{ R } =\frac{ 3 }{ R }\]
so if you have n parts it's
\[\frac{ n }{R }\]

Answer 20

yes, nice explanation , go on

Answer 21

we know that R is \[\frac{100 }{ n }\]

Answer 22

yes, then

Answer 23

so \[1=\frac{ n }{ R }=\frac{ n }{ \frac{ 100 }{ n } }=\frac{ n*n }{ 100 }\]
so \[1=\frac{ n^2 }{ 100 }\]
\[100=n^2\] so
n=10 because it cannot be -10

Answer 24

Do you unerstand it?

Answer 25

yeah, i understand,,, thank you.. @Fifciol

Answer 26

yw:)