A 12 V source feeds a network with a 4 Ω resistor from the top node to A. From A to the bottom node there are two 4 Ω branches, one directly and one through an ideal forward-biased diode to B and then 4 Ω to the bottom. Statements: (A) ammeter current is 2 A; (B) net resistance is 8 Ω; (C) voltage across BC is 4 V; (D) diode current is 1 A. Choose the correct option.
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Question

A 12 V source feeds a network with a 4 Ω resistor from the top node to A. From A to the bottom node there are two 4 Ω branches, one directly and one through an ideal forward-biased diode to B and then 4 Ω to the bottom. Statements: (A) ammeter current is 2 A; (B) net resistance is 8 Ω; (C) voltage across BC is 4 V; (D) diode current is 1 A. Choose the correct option.

Options

G1 Physics · Accepted Answer

The two lower 4 Ω branches are in parallel, giving 2 Ω; in series with upper 4 Ω gives R e q = 6 R_eq=6 R e ​ q = 6 Ω. Total current=12/6=2 A. It splits equally, so diode-branch current=1 A and V B C = 1 × 4 = 4 V_BC=1×4=4 V B ​ C = 1 × 4 = 4 V. Thus A,C,D are correct and B is false.