Grade 10Electric Current

A rod of resistance 4 ohm is turned into half circle.What is the resistance along the diameter

Maharani

6 Years agoGrade 10

2 Answers

Arun

6 Years ago

Let the resistance per unit length be $\lambda$ . Then,

$\lambda$ = 4/ $2\pi r$

=> resistance of the specimen wire connected along the diameter is:

$R = \lambda \times 2r = \frac{4}{\pi } \Omega$

Now after the specimen wire is connected along the diameter, three resistance come in parallel:

$2 \Omega$ (resistance of one half of the circumference), $2 \Omega$ (resistance of the other half of the circumference) and R(resistance of the diameter). Solving, we get:

${\color{DarkRed}R_{eq} = \frac{4}{4+\pi } \Omega }$

Rajat

6 Years ago

Perimeter of the half circle is πr + 2r where r is radius of the circle. Along the diameter the circle is divided into two unequal parts of length πr and 2r which are parallel to each other.

Resistance of each part:

1) 4*π/(π+2)

2) 4*2/(π+2)

Equivalent resistance is:

R1*R2/(R1+R2)

=[32π/(π+2)^2]/4

=8π/(π+2)^2