A) 2\[\Omega \]
B) 5\[\Omega \]
C) 7 a\[\Omega \]
D) 10\[\Omega \]
Correct Answer: B
Solution :
\[3\,\Omega \]resistor and\[7\,\Omega \]resistor are in series. Therefore resultant is\[=10\text{ }\Omega (7+3)\]. This\[10\text{ }\Omega \]equivalent resistance is in parallel with resistance\[(10\text{ }\Omega )\]in arm AC. \[\therefore \] \[\frac{1}{{{R}_{1}}}=\frac{1}{10}+\frac{1}{10}\] \[\Rightarrow \] \[{{R}_{1}}=5\,\Omega \] Now,\[{{R}_{1}}\]is in series with resistor\[(5\,\Omega )\]in arm CB. \[\therefore \] \[{{R}_{2}}=5+5=10\,\Omega \] Again\[{{R}_{2}}\]is in parallel with resistance\[(10\,\Omega )\]in arm AB \[\therefore \] \[{{R}_{2}}=5+5=10\,\Omega \] \[\Rightarrow \] \[R=5\,\Omega \]You need to login to perform this action.
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