question_answer

Four resistances 10\[\Omega \]  5 \[\Omega \] 7 \[\Omega \] and 3 \[\Omega \] are connected so that they form the sides of a rectangle AB, BC, CD, and DA respectively. Another resistance of 10 \[\Omega \] is connected across the diagonal AC. The equivalent resistance between A and B is

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 \[\therefore \] \[{{R}_{2}}=5+5=10\,\Omega \] \[\Rightarrow \] \[R=5\,\Omega \]