Question

Q.3 A cylindrical conductor of uniform cross-section has a resistance of 18 2. It is cut into three equal parts. If the three parts are connected in parallel, the effective resistance of the combination is 1. 20 2. 60 3. 120 4. 180​

Answer 1

Explanation:

A wire having a resistance of 18 Omega is cut into three equal parts. What is the equivalent resistance if these three parts are joined in parallel ? Since these three parts are joined in parallel, the equivalent resistance, R=r/3=6Ω/3=2Ω

Answer 2

Answer:

The effective resistance of the combination is 3Ω

Explanation:

We know that, resistance is directly proportional to the length. It means that when the length of the material increases, then the resistance of the material too increases.

So, when the cylindrical conductor is cut into 3 equal parts then the resistance of the cylindrical conductor gets divided into equal resistances.

Let us consider the resistance of each cut part on the whole to be R'. Therefore, the resistance of each cylindrical conductor would be,

R' = $R_{1} + R_{2} + R_{3}$

R' = 18/3

   = 6Ω

Therefore, the resistance of each equally cut cylindrical conductor is  6Ω.

If these 3 equally cut conductors are connected in parallel the the effective resistance could be calculated using the formula

$\frac{1}{R }$ = $\frac{1}{R_{1}} + \frac{1}{R_{2} } + \frac{1}{R_{3} }$

which could be written as

$\frac{1}{R}$ = $\frac{1}{6} + \frac{1}{6} + \frac{1}{6}$

$\frac{1}{R}$ = $\frac{3}{6}$

R = $\frac{6}{3}$

R = 2Ω

Therefore, the effective resistance of the combination is 2Ω.

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