Question

A uniform wire of resistance 9 ohm having resistance 1 ohm/m is bent in the form of a circle as shown in figure. If the equivalent resistance between P & Q is 2 ohm.what is the length of shorter section ?​

Answer 1

Therefore the length of shorter wire is 3 m.

Explanation:

Given:

Total resistance of the wire, $R=9\ \Omega$

resistivity of wire material, $\rho=1\ \Omega.m^{-1}$

equivalent resistance between points P&Q on the formed circle, $R_e=2\ \Omega$

Full length of wire:

$l=\frac{R}{\rho}$

$l=\frac{9}{1}$

$l=9\ m$

• Now, let the length of the shorter wire part be x.

Now according to the question the two parts of the wire will be in parallel connection:

$\frac{1}{R_e} =\frac{1}{R_s} +\frac{1}{R_l}$

where:

$R_s\ \&\ R_l$ are the resistance of shorter and longer wire respectively.

$\frac{1}{2} =\frac{1}{1(x)} +\frac{1}{1(9-x)}$

$x^2-9x+18=0$

$x=3\ or\ x=6$

Therefore the length of shorter wire is 3 m.

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Answer 2

The shorter length is 3 m.

Explanation:

Given that,

Resistance of wire = 9 ohm

Equivalent resistance = 2 ohm

Let resistance of shorter part AB be x .

The total resistance is 9Ω the resistance of longer AB part is (9−x) between point A and B the two resistance x.

We need to calculate the equivalent resistance

Using formula of equivalent resistance

$R_{eq}=\dfrac{(R-r)r}{R}$

Put the value into the formula

$2=\dfrac{(9-x)x}{9}$

$x^2-9x+18=0$

$x= 3, 6$

Hence, The shorter length is 3 m.

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Topic : shorter length

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