Question

please answer this question ​

Answer 1

Answer:

Explanation:

ANSWER:-

Given that:-

• Whole Resistance of wire = 40 ohm
• Now, it is cut in three equal lengths
• 3rd, joined in parallel.

Let's go step-by-step!

1. Let the first Resistance be R1
2. The second resistance be R2
3. And third one, R3.

According to Statement:-

$\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3} = \dfrac{1}{R_p}$

Now, R1, R2 and R3 are equal to $\dfrac{R}{3}$ as it is divided on 3 parts.

So, placing them , we get;-

$\dfrac{1}{\dfrac{R}{3}}+\dfrac{1}{\dfrac{R}{3}}+\dfrac{1}{\dfrac{R}{3}} = \dfrac{1}{R_p}$

$\dfrac{1}{R_p} = \dfrac{3}{R} + \dfrac{3}{R}+ \dfrac{3}{R}$

$\dfrac{1}{R_p} = \dfrac{9}{R}$

But, we know that the R is equal to 40!

[This is because the total Resistance of the wire is 40]

So, placing value of 40 in R,

$\dfrac{1}{R_p} = \dfrac{9}{40}$

$R_p = \dfrac{40}{9}$

4.44 $\Omega$ is the answer.

Note:-

• In parallel resistance, all things work independently.
• The formula is :-

$\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3} = \dfrac{1}{R_p}$

• R1, R2 and R3 are the other resistances.
• Here, R1=R2 = R3 as they are cut in equal parts.