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Q. The equivalent resistance across A and B isPhysics Question Image

JIPMERJIPMER 2010Current Electricity

Solution:

The equivalent circuit can be redrawn as
We have, $ \, \, \, \, \, \, \, \frac {P}{Q} = \frac {R}{S}$
ie, $\, \, \, \, \, \, \, \, \frac {4}{4} = \frac {4}{4}$
So, the given circuit is a balanced Wheatstone's bridge.
Hence, the equivalent resistance
$R_{AB} = \frac {(4 + 4 )\times (4 + 4 )}{(4 + 4 ) +(4 + 4 )}$
$\frac {8 \times 8}{8 + 8} = \frac {64}{16} = 4 \Omega$

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