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The current I drawn from the 5 volt source will be [Figure: a 5 V source between node L (positive) and node R (negative). L to M: 5 Ω; M to N: 10 Ω; L to N: 10 Ω (top); N to R: 20 Ω; M to R: 10 Ω (lower).]

Asked in JEE Main 2006 · Wheatstone bridge

Figure: Wheatstone bridge
Answer: (3) 0.5 A

Step-by-step solution

The network is a Wheatstone bridge with L-M 5 Ω, M-R 10 Ω, L-N 10 Ω, N-R 20 Ω and M-N 10 Ω as the bridge arm.

5/(10)=(10)/(20), so it is balanced and M-N carries no current.

R_eq=(5+10)∥(10+20)=(15×30)/(45)=10 Ω.

I=5/(10)=0.5 A.

Why the other options are wrong

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