Question

a metallic wire having a resistance of 20 ohm is bent in order to form a complete circle calculate the resistance between any two diametrically opposite points of the circle.

Answer 1

Answer:

ok

Explanation:

Answer 2

$R_{eq}=5\ \Omega$

Explanation:

Given:

resistance of metallic wire, $R=20\ \Omega$

Since the same wire is bent to form a circle so the length of the wire remains unchanged.

When we take resistance at any two points diametrically opposite to each other then the wire acts as two resistances in parallel.

We know relation between the resistance and the length of the wire is given as:

$R=\rho.\frac{l}{a}$ ..................................(1)

where:

$\rho=$ resistivity of the wire

$l=$ length of the wire

$a=$ area of the wire

Let the length of the wire be $l$ .

When we form the circle and choose two diametrically opposite points then we get a potential junction after the half length of the wire. The wire is electrically divided into the length $\frac{l}{2}$ as shown in the figure.

Now putting this value of length into the equation (1):

$R'=\rho .\frac{l}{2a}$

$R'=\frac{R}{2}$

$R'=10\Omega$ is the resistance of each wire segment.

Since these segments are in parallel:

$R_{eq}=\frac{R_1.R_2}{R_1+R_2}$

$R_{eq}=\frac{10\times 10}{10+10}$

$R_{eq}=5\ \Omega$

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TOPIC: resistance in parallel

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