The current $I$ drawn from the $5$ volt source will be ............... $A$
AIEEE 2006, Medium
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The network of resistors is a balanced wheatstone bridge. The equivalent circuit is
$R_{e q}=\frac{15 \times 30}{15+30}=10 \Omega \Rightarrow I=\frac{V}{R}=\frac{5}{10}=0.5 \mathrm{A}$
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