Question

Resistance of rod is $1\Omega$ . It is bent in the form of square. What is the resistance across adjoint corners ?

• A. $1\Omega$
• B. $3\Omega$
• C. $\frac{3}{16}\Omega$
• D. $\frac{3}{4}\Omega$

Answer 1

Answer: (C)

When rod is bent in the form of square, then each side has resistance of $\frac{1}{4}\Omega .$ As shown $R_1,R_2$ and $R_3$ are connected in series, so their equivalent resistance

$R=R_1+R_2+R_3$
$=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}$
$=\frac{3}{4}\Omega$
Now, R and $R_4$ are connected in parallel, so equivalent resistance of the circuit is
$R=\frac{R\times R_4}{R+R_4}$
$=\frac{(3/4)(1/4)}{(3/4)+(1/4)}$
$=\frac{3/16}{1}$
$=\frac{3}{16}\Omega$