A Thevenin equivalent with V_th = 9 V and R_th = 3 Ω feeds a 1 Ω load in series. What is the current through the load and the voltage across it?

A Thevenin equivalent with V_th = 9 V and R_th = 3 Ω feeds a 1 Ω load in series. What is the current through the load and the voltage across it?

• A. I = 2 A, V_load = 2 V
• B. I = 3 A, V_load = 3 V
• C. I = 2.25 A, V_load = 2.25 V
• D. I = 1 A, V_load = 1 V

Answer: (C)

In a Thevenin model, the internal resistance and the load sit in series, so the same current flows through both. The total resistance is 3 Ω + 1 Ω = 4 Ω, giving a loop current I = V_th / (R_th + R_load) = 9 V / 4 Ω = 2.25 A. The load, being in series, carries that same current, so V_load = I * R_load = 2.25 A * 1 Ω = 2.25 V. You can also see this as a voltage divider: V_load = V_th * (R_load / (R_th + R_load)) = 9 V * (1/4) = 2.25 V. The internal drop is V_Rth = I * R_th = 2.25 A * 3 Ω = 6.75 V, which together with V_load sums to 9 V.

In a Thevenin model, the internal resistance and the load sit in series, so the same current flows through both. The total resistance is 3 Ω + 1 Ω = 4 Ω, giving a loop current I = V_th / (R_th + R_load) = 9 V / 4 Ω = 2.25 A. The load, being in series, carries that same current, so V_load = I * R_load = 2.25 A * 1 Ω = 2.25 V. You can also see this as a voltage divider: V_load = V_th * (R_load / (R_th + R_load)) = 9 V * (1/4) = 2.25 V. The internal drop is V_Rth = I * R_th = 2.25 A * 3 Ω = 6.75 V, which together with V_load sums to 9 V.