A) \[100\,\Omega \]
B) \[180\,\Omega \]
C) \[220\,\Omega \]
D) \[440\,\Omega \]
Correct Answer: A
Solution :
Here: Resistance of arm \[PQ\text{ (}{{R}_{1}})=100\text{ }\Omega \] Resistance of arm \[QR\text{ (}{{R}_{2}})=100\text{ }\Omega \] Resistance of arm \[\text{RS (}{{R}_{3}})=100\text{ }\Omega \] Resistance of arm \[\text{SP (}{{R}_{4}})=100\text{ }\Omega \] Resistance of upper arm \[{{R}_{1}}\] and \[{{R}_{2}}\] are connected in series. Hence, their resultant resistance \[{{R}_{U}}={{R}_{1}}+{{R}_{2}}=100+100=200\,\Omega \] Similarly the resistance of lower half \[{{R}_{3}}\]and\[{{R}_{4}}\] are also connected in series. Hence, their resultant resistance is \[{{R}_{L}}={{R}_{3}}+{{R}_{4}}=100+100=200\,\Omega \] So, equivalent resistance \[\frac{1}{{{R}_{PR}}}=\frac{1}{{{R}_{U}}}+\frac{1}{{{R}_{L}}}\] \[=\frac{1}{200}+\frac{1}{200}=\frac{2}{200}=\frac{1}{100}\] So, \[{{R}_{PR}}=100\,\Omega \]
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