Question

A wire whose resistance is 80ohm is cut into three
Pieces of equal lengths which are then arranged in
Parallel. calculate the resistance of the combination

Answer 1

Given that, a wire whose resistance is 80 ohm is cut into three pieces of equal lengths.

We have to find the resistance of the combination. (means Req)

Assume a long cylindrical wire, which is cut into three equal pieces. When the wire of resistance 80 ohm cut into equal pieces then it's resistance = 80/3 ohm

Also, given that all the three pieces of wire are connected in parallel. So,

1/Rp = 1/R1 + 1/R2 + 1/R3

R1 = R2 = R3 = 80/3

Substitute value of R1, R2 and R3 in the above formula,

1/Rp = 1/(80/3) + 1/(80/3) + 1/(80/3)

1/Rp = 3/80 + 3/80 + 3/80

1/Rp = (3 + 3 + 3)/80

1/Rp = 9/80

Rp = 80/9

Rp = 8.89 ohm

Now,

Req = 1/R1 + 1/R2 + 1/R3

Req = (R1 + R2 + R3)/R1R2R3

Req = 1/Rp

Req = Rp

Req = 8.89 ohm

Therefore, the resistance of the combination is 8.89 ohm.

Answer 2

Answer:

${80\over9}~\Omega=8.\bar{88}~\Omega$

Explanation:

The resistance of a wire is given by

$R=\rho{l\over A}\\\\~~~~~~where,\\~~~~~~~~R=Resistance\\~~~~~~~~\rho=Resistivity\\~~~~~~~~l=length\\~~~~~~~~A=Area$

Note that, $\rho$ is constant for a given material of wire.

Let the initial length of wire be "$l$". It's cross section area "$A$" will be same, for the cut pieces too.

Hence, the initial resistance of wire

$R_1=80~\Omega$

And, for each of the cut pieces,

$R={80\over3}~\Omega~~~~~~~~Since~(R~is~directly~proportional~to~l)$

Hence, Equivalent resistance of three of them in parallel will be,

$R_{eq.}={{80\over3}\over3}\\\\R_{eq.}={80\over9}~\Omega$