A) \[2.5\,\Omega \]
B) \[7.5\,\Omega \]
C) \[10\,\Omega \]
D) \[12.5\,\,\Omega \]
Correct Answer: B
Solution :
Key Idea: The given circuit is a balanced Wheatstone bridge. 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 unaffected. 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) \[=\frac{15\times 15}{15+15}\] \[=\frac{15\times 15}{30}\] \[=7.5\,\Omega \]You need to login to perform this action.
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