The current flowing through R₂ is [Figure: an 8 V battery is connected between nodes E and A. R₅=2 Ω joins E to D. R₆=4 Ω joins E to A. R₇=3 Ω joins D to A. R₄=3 Ω joins D to C. R₃=6 Ω joins A to C. R₂=4 Ω joins C to B. R₁=2 Ω joins A to B.]

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Accepted Answer

(3) 1/3 A. R₆ sits directly across the 8 V battery, so it does not affect the rest. R₂+R₁=6 Ω is in parallel with R₃=6 Ω between C and A: 3 Ω. With R₄: 3+3=6 Ω from D to A, in parallel with R₇=3 Ω: 2 Ω. Add R₅: 2+2=4 Ω, so current from E is 8/4=2 A and V_DA=2×2=4 V. Current in R₄ branch =4/6=2/3 A, which splits equally between the two 6 Ω paths. I_R₂=1/3 A

The current flowing through R₂ is [Figure: an 8 V battery is connected between nodes E and A. R₅=2 Ω joins E to D. R₆=4 Ω joins E to A. R₇=3 Ω joins D to A. R₄=3 Ω joins D to C. R₃=6 Ω joins A to C. R₂=4 Ω joins C to B. R₁=2 Ω joins A to B.]

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