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The current I drawn from the 5 volt source will be

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0.5A

0.33A

0.67A

0.17 A

answer is A.

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Detailed Solution

The given circuit is a balanced Wheatstone’s network
Hence effective resistance $R = \frac{(10 + 20) (5 + 10)}{10 + 20 + 5 + 10}$
$= \frac{30 \times 15}{45} = 10 Ω ∴ i = \frac{V}{R} = \frac{5}{10} = 0.5 A$

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