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

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

Grade:10

2 Answers

Arun
25763 Points
one year 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
213 Points
one year 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
 

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