Q. Resistance of rod is 1 $ \Omega $ . It is bent in form of square. What is resistance across adjoint corners?
ManipalManipal 2008Current Electricity
Solution:
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 $
