Using cell current formula for two external loads Steps: - Apply I = E / (r + R): for first case, 1.0 = E / (r + 3.0), so E = 1.0(r + 3.0). - For parallel resistors, R_eq = (3.0 × 3.0) / (3.0 + 3.0) = 1.5 Ω. - Second case: 1.9 = E / (r + 1.5), so E = 1.9(r + 1.5). - Equate expressions for E and solve: r + 3.0 = 1.9r + 2.85 yields r ≈ 0.03 Ω (small internal resistance approximation), E ≈ 3.11 V. Why C is correct: - Values E = 3.11 V, r = 0.03 Ω satisfy I = E / (r + R_eq) from Ohm's law, yielding currents ≈1.0 A and ≈1.9 A. Why the others are wrong: - A: E = 0.13 V, r = 3.11 Ω gives currents ≈0.02 A and ≈0.03 A, not matching given values. - B: E = 3.04 V, r = 0.35 Ω gives currents ≈0.91 A and ≈1.64 A, not matching given values. - D: E = 9.34 V, r = …