JEE Main Physics Class 12 Current Electricity Part 2 Questions

Let the sides of square be a.
∴ Total length or perìmeter of square =4a
If the radius of shape of circle be r, then
∴2πr=4a
⇒r=

4a

2π

⇒r=

2a

π

Since, resistance of each side of square = 3 Ω
∴ Total resistance of square =4×3=12Ω
i.e. resistance of length 2πr=12Ω
⇒ Resistance of πr=6Ω Now, equivalent resistance of circle diametrically opposite
(Req)=

R1R2

R1+R2

=

6×6

6+6

=3Ω