Transform a 10 V voltage source in series with a 5 Ω resistor to its Norton equivalent.

Transform a 10 V voltage source in series with a 5 Ω resistor to its Norton equivalent.

• A. I_N = 2 A; R_N = 5 Ω
• B. I_N = 1 A; R_N = 5 Ω
• C. I_N = 2 A; R_N = 10 Ω
• D. I_N = 4 A; R_N = 5 Ω

Answer: (A)

Transforming a voltage source with a series resistor into a Norton equivalent uses I_N = V_s / R_s and R_N = R_s. The 10 V source in series with 5 Ω becomes a 2 A current source in parallel with 5 Ω. This matches the short-circuit current (10 V / 5 Ω = 2 A) and leaves the same impedance seen from the terminals, 5 Ω. The Norton form preserves the same behavior for any load connected across the terminals. So the correct Norton equivalent is a 2 A current source in parallel with 5 Ω.

Transforming a voltage source with a series resistor into a Norton equivalent uses I_N = V_s / R_s and R_N = R_s. The 10 V source in series with 5 Ω becomes a 2 A current source in parallel with 5 Ω. This matches the short-circuit current (10 V / 5 Ω = 2 A) and leaves the same impedance seen from the terminals, 5 Ω. The Norton form preserves the same behavior for any load connected across the terminals. So the correct Norton equivalent is a 2 A current source in parallel with 5 Ω.