Question

6. A wire of resistance 20ohm is bent in the form of closed
circle. What is the effective resistance between two
points at the ends of any diameter of the circle ?​

Answer 1

We have:-

• A wire of resistance 20ohm is bent in the form of closed circle.

To Determine:-

• Effective resistance between two points at the ends of any diameter of the circle?

Let's do it then:

We know that, Resistance of any conductor is directly proportional to its length. Here, the R will vary with L because Cross-sectional area and Resistivity will remain same.

Hence:

The diametrically opposite points divides the resistance into half of the total resistance.

• R1 = R2 = 20 ohms / 2 = 10 ohms

Now:

We have two resistances connected in parallel. Then equivalent resistance is:

$\sf{R_{eq. }= \dfrac{R1R2}{R1 + R2} }$

Plugging the resistances,

⇒ Req. = 10 × 10 / 10 + 10 ohms

⇒ Req. = 100 / 20 ohms

⇒ Req. = 5 ohms.

Therefore:-

• Effective resistance between two points at the ends of any diameter of the circle is 5 ohms.

Answer 2

Answer:

$\huge \bf Answer$

• Resistance of wire = 20 ohm
• R1 = R2 = 20 ohms / 2 = 10 ohms

Now,

As we know that resistance of any conductor is directly proportional on its length.

Hence,

$\huge \bf {R}_{eq} = \dfrac {R1R2}{R1+R2}$

$\sf \implies \: req \: = \dfrac{10 \times 10}{10 + 10}$

$\sf \implies \: req \: = \dfrac{100}{20}$

$\huge \bf \: req \: = 5 \: ohms$

Therefore

Effective resistance between two points at the ends of any diameter of the circle is 5 ohms.