A wire is bent in the form of a triangle now the equivalent resistance R between its one end and the midpoint of the side is

Question

A wire is bent in the form of a triangle now the equivalent resistance R between its one end and the midpoint of the side is (A)  (5R/12)  (B)  (7R/12)  (C)  (3R/12)  (D)  (R/12)

Tardigrade Admin · Accepted Answer

Resistances R, R and (R/2) are in series, their resultant resistance, R=R+R+(R/2)=(5R/2) Resistances R and (R/2) are in parallel. Their resultant resistance (1/R)=(2/5R)+(2/R)=(2+10/5R) R=(5R/12)

Q. A wire is bent in the form of a triangle now the equivalent resistance R between its one end and the midpoint of the side is

$\frac{5 R}{12}$

$\frac{7 R}{12}$

$\frac{3 R}{12}$

$\frac{R}{12}$

Resistances $R, R$ and $\frac{R}{2}$ are in series, their resultant resistance, $R = R + R + \frac{R}{2} = \frac{5 R}{2}$ Resistances $R$ and $\frac{R}{2}$ are in parallel. Their resultant resistance $\frac{1}{R} = \frac{2}{5 R} + \frac{2}{R} = \frac{2 + 10}{5 R}$ $R = \frac{5 R}{12}$

Q. A wire is bent in the form of a triangle now the equivalent resistance R between its one end and the midpoint of the side is

125R​

127R​

123R​

12R​

Resistances R,R and 2R​ are in series, their resultant resistance, R=R+R+2R​=25R​ Resistances R and 2R​ are in parallel. Their resultant resistance R1​=5R2​+R2​=5R2+10​ R=125R​

$\frac{5 R}{12}$

$\frac{7 R}{12}$

$\frac{3 R}{12}$

$\frac{R}{12}$

Resistances $R, R$ and $\frac{R}{2}$ are in series, their resultant resistance, $R = R + R + \frac{R}{2} = \frac{5 R}{2}$ Resistances $R$ and $\frac{R}{2}$ are in parallel. Their resultant resistance $\frac{1}{R} = \frac{2}{5 R} + \frac{2}{R} = \frac{2 + 10}{5 R}$ $R = \frac{5 R}{12}$