The current I drawn from the 5 volt source will be (a) 0.33 A (b) 0.5 A (c) 0.67 A (d) 0.17 A

Answer: (b). Label the node after 5 Ω as Y and the node after the middle 10 Ω as Z. Arms: 5 Ω and 10 Ω from the + end, 10 Ω and 20 Ω to the − end; 5/10 = 10/20, so the bridge is balanced and the middle 10 Ω (Y to Z) carries no current. Paths: 5 + 10 = 15 Ω and 10 + 20 = 30 Ω in parallel: R = 10 Ω. I = 5/10 = 0.5 A So the answer is (b) 0.5 A.

The current I drawn from the 5 volt source will be | JEE Main 2006 (AIEEE)

Q196JEE Main 2006 · AIEEEMedium

Kirchhoff's Laws & Measuring Instruments

The current $I$ drawn from the $5$ volt source will be

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Answer: (b)

Label the node after $5\ \Omega$ as Y and the node after the middle $10\ \Omega$ as Z.
Arms: $5\ \Omega$ and $10\ \Omega$ from the + end, $10\ \Omega$ and $20\ \Omega$ to the − end; $\frac{5}{10} = \frac{10}{20}$ , so the bridge is balanced and the middle $10\ \Omega$ (Y to Z) carries no current.
Paths: $5 + 10 = 15\ \Omega$ and $10 + 20 = 30\ \Omega$ in parallel: $R = 10\ \Omega$ .
$I = \frac{5}{10} = 0.5\text{ A}$
So the answer is (b) 0.5 A.

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