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Q. In the following network, the current flowing through 15 $\Omega$ resistance is
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MHT CETMHT CET 2018Current Electricity

Solution:

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Above circuit is a balanced Wheatstone bridge.
Hence, the above circuit is drawn in the following manner
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From the above circuit,
$18 I =24(21-1) $
$18 I =504-24I$
$24 I+18I =504 $
$42I=504 $
$I =\frac{504}{42} $
$ \therefore I =1.2 A $