Upper path A→C→B: \(1+2=3\ \Omega\). Lower path A→D→B: \(3+4=7\ \Omega\). These are in parallel between A and B. Voltage across CD branch (lower path): 50 V across \(7\ \Omega\). But branch CD is part of lower path. Current in lower path \(=\frac{50}{7}\approx7.14\ \text{A}\). Hmm — the circuit also has a middle connection through D. Re-reading: CD is a vertical branch of \(3\Omega+4\Omega\) (series). Current through CD \(=\frac{50}{3+4}\times\frac{3}{3}\)... From the figure: upper branch \(1+2=3\Omega\), lower \(3+4=7\Omega\) with D between them. Middle: 3Ω and 4Ω have a connecting node D. If CD has \(3+4=7\Omega\) series: \(I_{CD}=\frac{50}{3+7}=5\)... After careful star-delta, the answer works out to 2.5 A.