question_answer

Four resistances 100, 50, 70 and 30 are connected so that they form the sides of a rectangle AB, BC, CD, and DA respectively. Another resistance of 100 is connected across the diagonal AC. The equivalent resistance between A and B is

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 \[\therefore \] \[\frac{1}{R}=\frac{1}{10}+\frac{1}{10}\]\[\Rightarrow \]\[R=5\Omega \]