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 0 is connected across the diagonal AC. The equivalent resistance between A and B is

A)  \[2\,\Omega \]                               

B)  \[5\,\Omega \]

C)  \[7\,\Omega \]                               

D)  \[10\,\Omega \]

Correct Answer: B

Solution :

                 \[3\,\Omega \]resistor and\[7\,\Omega \]resistor are in series. Therefore there 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 GB. \[\therefore \]  \[{{R}_{2}}=5+5=10\,\Omega \] Again\[{{R}_{2}}\]is in parallel with resistance\[(10\,\Omega )\]in arm AB \[\therefore \]  \[\frac{1}{R}=\frac{1}{10}+\frac{1}{10}\] \[\Rightarrow \]               \[R=5\,\Omega \]