#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

Arun
25763 Points
one year ago

Let the resistance per unit length be $\dpi{80} \lambda$. Then,
$\dpi{80} \lambda$ = 4/$\dpi{80} 2\pi r$
=> resistance of the specimen wire connected along the diameter is:
$\dpi{80} R = \lambda \times 2r = \frac{4}{\pi } \Omega$
Now after the specimen wire is connected along the diameter, three resistance come in parallel:
$\dpi{80} 2 \Omega$ (resistance of one half of the circumference),  $\dpi{80} 2 \Omega$ (resistance of the other half of the circumference) and R(resistance of the diameter). Solving, we get:
$\dpi{80} {\color{DarkRed}R_{eq} = \frac{4}{4+\pi } \Omega }$
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