Current division in parallel branches with series resistors

Steps:

• The circuit has three parallel branches from the power source: one with 1 resistor (A3), one with 2 resistors (A1), and one with 3 resistors (A4); A2 measures total current.
• Voltage V is the same across branches; current I = V / (nR), where n is number of resistors, so I ∝ 1/n.
• Given A1 = 0.6 A in 2R branch, V = 0.6 × 2R = 1.2R.
• A3 (1R branch) = 1.2R / R = 1.2 A; A4 (3R branch) = 1.2R / (3R) = 0.4 A.
• Total A2 = 0.6 + 1.2 + 0.4 = 2.2 A.

Why D is correct:

• Currents are inversely proportional to branch resistances (Ohm's law), yielding exact integer resistor counts and KCL balance.

Why the others are wrong:

• A: Total 1.0 A ≠ 0.6 + 0.6 + 0.2 = 1.4 A (violates KCL).
• B: Currents 0.6:0.3:0.1 yield non-integer resistor ratios (1.67:3.33:10).
• C: Currents 0.6:0.9:0.3 yield non-integer resistor ratios (1.5:1:3).

Final answer: D