Question

The ratio of equivalent resistance across A and B when switch is open to that when switch is closed is:

• A. $\frac{3}{8}$
• B. $\frac{5}{8}$
• C. $\frac{7}{8}$
• D. $\frac{9}{8}$

Answer 1

The correct option is D $\frac{9}{8}$

When switch is open :

$6\Omega$ and $12\Omega$ are connected in series and $12\Omega$ and $6\Omega$ are also connected in series and both these are in parallel connection.

So, $R_{AB}=(6+12)\parallel (12+6)$

$⟹R_{AB}=18\parallel 18=9\Omega$

When switch is closed :

$6\Omega$ and $12\Omega$ are connected in parallel and $12\Omega$ and $6\Omega$ are also connected in parallel and both these are in series connection.

So, $R_{AB}^{\prime }=[6\parallel 12]+[12\parallel 6]$

Or, $6\parallel 12=\frac{6\times 12}{6+12}=4\Omega$

Or, $R_{AB}^{\prime }=4+4=8\Omega$

Ratio $\frac{R_{AB}}{R_{AB}^{\prime }}=\frac{9}{8}$