A) \[2.5\,\,\Omega \]
B) \[7.5\,\,\Omega \]
C) \[10\,\,\Omega \]
D) \[12.5\,\,\Omega \]
Correct Answer: B
Solution :
We can show the network as below From the circuit \[\frac{10}{5}\,=\frac{10}{5}\] i.e., 2 = 2 So, it is balanced Wheatstone's bridge. Therefore, resistance of its middle arm will remain behaves as open circuited (inactive). So the net resistance in upper arms \[{{R}_{U}}=10+5=15\,\Omega \] (series) The net resistance in lower arms \[{{R}_{L}}=10+5=15\,\Omega \] (series) Hence, equivalent resistance of the network \[R=\frac{{{R}_{U}}\times {{R}_{L}}}{{{R}_{U}}+{{R}_{L}}}\] (parallel) \[=\frac{15\times 15}{15+15}=7.5\,\,\Omega \]You need to login to perform this action.
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