Which is the Norton equivalent of a 10 V source in series with a 5 Ω resistor?

Which is the Norton equivalent of a 10 V source in series with a 5 Ω resistor?

• 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 = 0 A; R_N = 5 Ω

Answer: (A)

In Norton form, a voltage source in series with a resistor becomes a current source in parallel with the same resistor. The parallel resistance equals the original series resistor, so R_N = 5 Ω. The Norton current is the short-circuit current at the output, which is V_th / R_th = 10 V / 5 Ω = 2 A. Therefore, the Norton equivalent is a 2 A current source in parallel with 5 Ω. This matches the described Norton values I_N = 2 A and R_N = 5 Ω.

In Norton form, a voltage source in series with a resistor becomes a current source in parallel with the same resistor. The parallel resistance equals the original series resistor, so R_N = 5 Ω. The Norton current is the short-circuit current at the output, which is V_th / R_th = 10 V / 5 Ω = 2 A. Therefore, the Norton equivalent is a 2 A current source in parallel with 5 Ω. This matches the described Norton values I_N = 2 A and R_N = 5 Ω.