In the given circuit diagram, a wire is joining points B and D. The current in this wire is [Figure: upper path A to B through 1 Ω, then B to C through 2 Ω. Lower path A to D through 4 Ω, then D to C through 3 Ω. A plain wire joins B and D. A 20 V ideal battery is connected between A and C.]

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(3) 2 A. With B and D joined: 1∥4=0.8 Ω and 2∥3=1.2 Ω, in series: 2 Ω. Total current =20/2=10 A. V_AB=10×0.8=8 V: current in 1 Ω is 8 A, in 4 Ω is 2 A. V_BC=10×1.2=12 V: current in 2 Ω is 6 A, in 3 Ω is 4 A. At B: 8 A in, 6 A out, so 2 A flows from B to D.

In the given circuit diagram, a wire is joining points B and D. The current in this wire is [Figure: upper path A to B through 1 Ω, then B to C through 2 Ω. Lower path A to D through 4 Ω, then D to C through 3 Ω. A plain wire joins B and D. A 20 V ideal battery is connected between A and C.]

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