A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is: (E is mid-point of arm CD)

Question

A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is: (E is mid-point of arm CD) (A) R (B) (1/16)R (C) (7/64)R (D) (3/4)R

Tardigrade Admin · Accepted Answer

(1/Req) = (8/7R) + (8/R) (1/req) = (8+56/7R) Req = (7R/64)

Q. A wire of resistance $R$ is bent to form a square $A BC D$ as shown in the figure. The effective resistance between $E$ and $C$ is : ( $E$ is mid-point of arm $C D$ )

$R$

$\frac{1}{16} R$

$\frac{7}{64} R$

$\frac{3}{4} R$

$\frac{1}{R _{e q}} = \frac{8}{7 R} + \frac{8}{R}$
$\frac{1}{r _{e q}} = \frac{8 + 56}{7 R}$
$R_{e q} = \frac{7 R}{64}$

Q. A wire of resistance R is bent to form a square ABCD as shown in the figure. The effective resistance between E and C is : (E is mid-point of arm CD)

R

161​R

647​R

43​R

Req​1​=7R8​+R8​ req​1​=7R8+56​ Req​=647R​

Q. A wire of resistance $R$ is bent to form a square $A BC D$ as shown in the figure. The effective resistance between $E$ and $C$ is : ( $E$ is mid-point of arm $C D$ )

$R$

$\frac{1}{16} R$

$\frac{7}{64} R$

$\frac{3}{4} R$

$\frac{1}{R _{e q}} = \frac{8}{7 R} + \frac{8}{R}$
$\frac{1}{r _{e q}} = \frac{8 + 56}{7 R}$
$R_{e q} = \frac{7 R}{64}$