Power dissipation determines current and total resistance

Steps:

• Use P = I²R for fixed resistor: I = √(P/R) = √(4/20) = √(1/5) A.
• Total resistance R_total = V / I = 16 / √(1/5) = 16√5 Ω ≈ 35.8 Ω.
• Variable resistance R_var = R_total - 20 Ω ≈ 35.8 - 20 = 15.8 Ω ≈ 16 Ω.
• Select closest option: 16 Ω.

Why A is correct:

• Matches calculation from P = I²R (Ohm's law for power) and V = I R_total, yielding R_var ≈ 16 Ω.

Why the others are wrong:

• B: Gives R_total = 56 Ω, I ≈ 0.286 A, P ≈ 1.63 W (too low).
• C: Gives R_total = 64 Ω, I ≈ 0.25 A, P ≈ 1.25 W (too low).
• D: Gives R_total = 84 Ω, I ≈ 0.19 A, P ≈ 0.72 W (too low).

Final answer: A