A)
10/3\[\Omega \]
B) 20/3\[\Omega \]
C) 10/5\[\Omega \]
D) 5\[\Omega \]
Correct Answer: A
Solution :
\[\because \] \[\frac{2}{3}=\frac{4}{6}\] \[\therefore \] Given, circuit is a balanced Wheatstones Bridge. So, 10 Q, resistance will be ineffective. Equivalent resistance of upper arms \[{{R}_{1}}=2+3=5\,\Omega \] \[\therefore \] Equivalent resistance of lower arms \[{{R}_{2}}=4+6=10\,\Omega \] \[{{R}_{AB}}=\frac{{{R}_{1}}{{R}_{2}}}{{{R}_{1}}+{{R}_{2}}}\] \[=\frac{5\times 10}{5+10}\] \[=\frac{50}{10}=\frac{10}{3}\Omega \]
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