Question

A wire of 10 ohm is bemt into a circle calculate the effective resistance between diameter a and b

Answer 1

Solution :

✴ Correct question :-

▪ A wire of 10Ω is bent into a circle, calculate the effective resistance between points A and B. (See the attachment for better understanding)

✴ Calculation :-

✒ Resistance per unit length of circle,

• $\sf\:\lambda=\dfrac{R}{2\pi{r}}$

✒ Length of sections ADB and ACB are r$\Theta$ and r$(2\pi-\Theta)$

• Resistance of section ADB

$\sf\:R_1=\lambda{r}\Theta=\dfrac{R\Theta}{2\pi}$

• Resistance of section ACB

$\sf\:R_2=\lambda{r}(2\pi-\Theta)=\dfrac{R(2\pi-\Theta)}{2\pi}$

✒ Now, R1 and R2 are connected in parallel between A and B then

$\sf\:R_{eq}=\dfrac{R_1R_2}{R_1+R_2}=\red{\dfrac{R\theta(2\pi-\theta)}{4\pi^2}}$

✏ Putting angle = 180° = π and R = 10Ω

$\sf\:Req=\dfrac{10\pi(2\pi-\pi)}{4\pi^2}\\ \\ \sf\: Req=\dfrac{10}{4}\\ \\ \bigstar\:\boxed{\mathtt{\purple{\large{R_{eq}=2.5\:\Omega}}}}$

Answer 2

Answer:

A wire has been bent into a circle . we have to calculate the effective resistance beyween terminals A and B

First of all, we should understand the following relationship :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{R = \rho \: (\dfrac{l}{a} )}}$

Keeping the area of cross-section constant , we can say that :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \bold{R \: \propto \: l}}$

Now the terminals A and B divide the circular structure into 2 semi-circles. And we can also say that each half resistance is in parallel combination with the other.

Resistance of each half of circle will be :

$\bigstar \: \: \: \sf{R1 = R2 = \dfrac{10}{2} = 5 \: ohm}$

Now R1 and R2 are in parallel with each other .

$\therefore \dfrac{1}{R \: eq.} = \dfrac{1}{5} + \dfrac{1}{5}$

$\implies \dfrac{1}{R \: eq.} = \dfrac{2}{5}$

$\implies R \: eq.= \dfrac{5}{2} ohm$

$\implies R \: eq.= 2.5 \: ohm$

So final answer :

$\boxed{ \blue{ \huge{ \bold{R \: eq.= 2.5 \: ohm}}}}$