question_answer

Four resistance \[\text{1}0\Omega ,\text{5}\Omega ,\text{7}\Omega \] and\[3\Omega \]. Are connected so that they form the side 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) \[7\Omega \]                                    

D) \[10\Omega \]

Correct Answer: B

Solution :

\[3\Omega \] resister and \[7\Omega \] resister are in series. Therefore, resultant \[=\text{1}0\Omega =\left( \text{7}+\text{3} \right)\] This \[\text{1}0\Omega \] equivalent resistance is in parallel with resistance (\[\text{1}0\Omega \]) in are AC \[\frac{1}{{{R}_{1}}}=\frac{1}{10}+\frac{1}{10}\] \[{{R}_{1}}=5\Omega \] Now. \[{{R}_{1}}\] is in series with resister \[5\Omega \] in are CB \[{{R}_{2}}=5+5\] \[R=10\Omega \] Again. \[{{R}_{2}}\] is in parallel with resistance (\[10\Omega \]) in are .AB \[\frac{1}{R}=\frac{1}{10}+\frac{1}{10}\] \[R=5\Omega \]