Question

A wire of resistance 5ohm is charged into a form of circle . What is the effective resistance between diameterlly two opposite of the circular wire .

Answer 1

Answer:

You should remember the fact that keeping physical quantities and area of cross section constant,

the resistance is directly proportional to the length of the conductor

Now, as per the question, the wire is made into that circle.

If we divide the circle in 2 opposite diametrically points , the wire is divided into 2 wires having half length.

So each half is 5/2 ohms.

The wires are parallel in connection. please refer to the attached photo to understand better.

So equivalent resistance is 5/4 ohms

Answer 2

$\bold{\underline{\underline{Answer:}}}$

Effective resistance = $\bold{\dfrac{5}{4}}$ ohm

$\bold{\underline{\underline{Step\:by\:-\:step\:explanation:}}}$

• In the above question we have been provided with a wire of resistance 5 ohm which is charged (formed) into a circle.

To find :

• We are asked to calculate the effective resistance between diameterally two opposite ends of the circular framed wire.

Solution :

The wire having resistance 5 ohm is formed into a circle. Hence we can imagine it to be formed into two semis.

Resistance = 5 ohm [Given]

Resistance of each semi = $\bold{\dfrac{5}{2}}$

$\bold{Resistance\:of\:semi_1}$ = AOB = $\bold{\dfrac{5}{2}}$

$\bold{Resistance\:of\:semi_2}$ = APB = $\bold{\dfrac{5}{2}}$

Now, to get the effective resistance we will sum up the resistance of each semi i.e semi 1 and semi 2.

Let R be the effective resistance.

Let $\bold{R_1}$ be the resistance of semi AOB.

Let $\bold{R_2}$ be the resistance of semi APB.

$\rightarrow$$\bold{\dfrac{1}{R}}$ = $\bold{\dfrac{1}{R_1}}$ + $\bold{\dfrac{1}{R_2}}$

$\rightarrow$$\bold{\dfrac{1}{R}}$ = $\bold{\dfrac{1}{5/2}}$ + $\bold{\dfrac{1}{5/2}}$

$\rightarrow$$\bold{\dfrac{1}{R}}$ = $\bold{\dfrac{2}{5}}$ + $\bold{\dfrac{2}{5}}$

$\rightarrow$$\bold{\dfrac{1}{R}}$ = $\bold{\dfrac{2+2}{5}}$

$\rightarrow$$\bold{\dfrac{1}{R}}$ = $\bold{\dfrac{4}{5}}$

Via reciprocal,

$\rightarrow$$\bold{R=\:{\dfrac{5}{4}}}$

•°• Effective resistance, R = $\bold{\frac{5}{4}}$ ohm.