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Asked in NEET 2025 · Kirchhoff's laws in networks
The wire CD puts C and D at one potential, so 1 Ω∥3 Ω is in series with 2 Ω∥4 Ω.
R_AB=(1×3)/4+(2×4)/6=3/4+4/3=(25)/(12) Ω.
I=(50)/(25/12)=24 A.
Left pair shares 24 A inversely as resistance: 1 Ω carries 18 A, 3 Ω carries 6 A.
Right pair: 2 Ω carries 16 A, 4 Ω carries 8 A.
At C: 18 A arrives and 16 A leaves through 2 Ω, so 2 A flows from C to D.
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