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Q.
Resistance of rod is $ 1\Omega $ . It is bent in the form of square. What is the resistance across adjoint corners ?
Bihar CECEBihar CECE 2001Current 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 $
