A) 20
B) 50
C) 7 ft
D) 100
Correct Answer: B
Solution :
\[3\Omega \]resistor and \[7\Omega \] resistor are in series. Therefore resultant is \[=10\Omega (7+3).\] This \[10\,\Omega \] equivalent resistance is in parallel with resistance \[(10\,\Omega )\] 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 \[\therefore \] \[\frac{1}{R}=\frac{1}{10}+\frac{1}{10}\]\[\Rightarrow \]\[R=5\Omega \]You need to login to perform this action.
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