Question

A wire of length l and resistance R is bent to form a circle. What is the

effective resistance between two points along its diameter ?​

Answer 1

Let's find the attachment for the given scenario

Resistance of the wire (R) = R Ω

Effective resistance along diameter AB is given by ,

R/2 , R/2 are connected in parallel ,

$\to \sf \dfrac{1}{R_{ef}}=\dfrac{1}{\frac{R}{2}}+\dfrac{1}{\frac{R}{2}}\\\\\to \sf \dfrac{1}{R_{ef}}=\dfrac{2}{R}+\dfrac{2}{R}\\\\\to \sf \dfrac{1}{R_{ef}}=\dfrac{4}{R}\\\\\leadsto \sf \pink{R_{ef}=\dfrac{R}{4}}\ \; \bigstar$

Effective resistance along the diameter is R/4

More Info :

If resistors are connected in series ,

$\sf \bigstar\ \; R_{eq}=R_1+R_2+R_3+...$

If resistors are connected in parallel ,

$\sf \bigstar\ \; \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...$

Answer 2

Answer:

Resistance of the wire (R) = R Ω

Effective resistance along diameter AB is given by ,

R/2 , R/2 are connected in parallel ,

\begin{gathered}\to \sf \dfrac{1}{R_{ef}}=\dfrac{1}{\frac{R}{2}}+\dfrac{1}{\frac{R}{2}}\\\\\to \sf \dfrac{1}{R_{ef}}=\dfrac{2}{R}+\dfrac{2}{R}\\\\\to \sf \dfrac{1}{R_{ef}}=\dfrac{4}{R}\\\\\leadsto \sf \pink{R_{ef}=\dfrac{R}{4}}\ \; \bigstar\end{gathered}

→

R

ef

1

=

2

R

1

+

2

R

1

→

R

ef

1

=

R

2

+

R

2

→

R

ef

1

=

R

4

⇝R

ef

=

4

R

★

Effective resistance along the diameter is R/4

More Info :

If resistors are connected in series ,

\sf \bigstar\ \; R_{eq}=R_1+R_2+R_3+...★ R

eq

=R

1

+R

2

+R

3

+...

If resistors are connected in parallel ,

\sf \bigstar\ \; \dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+...★

R

eq

1

=

R

1

1

+

R

2

1

+

R

3

1

+...