Question

Hi Friends !!
A wire of 10Ω is bent into a circle, calculate the effective resistance between points A and B

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Answer 1

Aɴꜱᴡᴇʀ

${\mathtt{\green{\large{R_{eq}=2.5\:\Omega}}}}$

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Gɪᴠᴇɴ

A wire of 10Ω is bent into a circle

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ᴛᴏ ꜰɪɴᴅ

Effective resistance between A and B

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Sᴛᴇᴘꜱ

• Resistance per unit length of circle,

$\large \green{\tt\:\lambda=\dfrac{R}{2\pi{r}}}$

• Length of sections ADB and ACB are r\ThetaΘ and r(2\pi-\Theta)(2π−Θ)

• Resistance of section ADB

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

• Resistance of section ACB

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

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

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

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

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

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$p

Answer 2

Explanation:

The resistance across a diameter of the circle is 2.5 ohms.

Explanation:

It is given that, a wire of resistance 10 ohm is bent to form a close the circle. We need to find the resistance across a diameter of the circle.

When it is bent, the ends of the diameter are considered then the resistor will get split up into and they will be now in parallel. The equivalent is given by :

\begin{gathered}\dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{5}\\\\R=2.5\ \Omega\end{gathered}

R

1

=

5

1

+

5

1

R=2.5 Ω

So, the resistance across a diameter of the circle is 2.5 ohms. Hence, this is the required solution

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