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Q.
In the circuit shown the equivalent resistance between $A$ and $B$ is
Chhattisgarh PMTChhattisgarh PMT 2009
Solution:
The three resistances between $A$ and $B$ are parallel,
$\frac{1}{R_{\text {comb }}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$
$=\frac{1}{9}+\frac{1}{9}+\frac{1}{9} \frac{1}{R_{\text {comb }}}=\frac{3}{9} $
$\Rightarrow R_{\text {comb }}=3\, \Omega$