For a 6 V source in series with 4 Ω, determine V_th, R_th, I_N, and R_N.

For a 6 V source in series with 4 Ω, determine V_th, R_th, I_N, and R_N.

• A. V_th = 6 V; R_th = 4 Ω; I_N = 1.5 A; R_N = 4 Ω
• B. V_th = 6 V; R_th = 2 Ω; I_N = 3 A; R_N = 2 Ω
• C. V_th = 4 V; R_th = 6 Ω; I_N = 1 A; R_N = 6 Ω
• D. V_th = 6 V; R_th = 4 Ω; I_N = 1 A; R_N = 8 Ω

Answer: (A)

This question tests Thevenin/Norton equivalence for a simple two-terminal network. With a 6 V source in series with a 4 Ω resistor, the open-circuit voltage across the output terminals is just the source voltage, so V_th = 6 V. To get the Thevenin resistance, turn off the independent source (replace the 6 V source with a short circuit); the resistance seen from the terminals is then just 4 Ω, so R_th = 4 Ω. The Norton current is the short-circuit current between the terminals; shorting the output makes the current 6 V divided by 4 Ω, giving I_N = 1.5 A. The Norton resistance R_N equals R_th (and also V_th / I_N), so R_N = 4 Ω.

This question tests Thevenin/Norton equivalence for a simple two-terminal network. With a 6 V source in series with a 4 Ω resistor, the open-circuit voltage across the output terminals is just the source voltage, so V_th = 6 V. To get the Thevenin resistance, turn off the independent source (replace the 6 V source with a short circuit); the resistance seen from the terminals is then just 4 Ω, so R_th = 4 Ω. The Norton current is the short-circuit current between the terminals; shorting the output makes the current 6 V divided by 4 Ω, giving I_N = 1.5 A. The Norton resistance R_N equals R_th (and also V_th / I_N), so R_N = 4 Ω.