A Thevenin circuit has V_th = 24 V, R_th = 6 Ω. A load of 4 Ω is connected in series with the Thevenin source. What is the current through the load and the voltage across it?

A Thevenin circuit has V_th = 24 V, R_th = 6 Ω. A load of 4 Ω is connected in series with the Thevenin source. What is the current through the load and the voltage across it?

• A. I = 24/6 = 4 A; V_L = 24 V
• B. I = 24/(6+4) = 2.4 A; V_L = I*4 = 9.6 V
• C. I = 24/4 = 6 A; V_L = 24/6 = 4 V
• D. I = 24/(6-4) = 12 A; V_L = 48 V

Answer: (B)

Key idea: in a Thevenin circuit, the load sits in series with the internal resistance, so the same current flows through both and is set by the total resistance. The current through the load is I = V_th / (R_th + R_L) = 24 / (6 + 4) = 2.4 A. The voltage across the load is V_L = I × R_L = 2.4 A × 4 Ω = 9.6 V. You can also see this as a voltage divider: V_L = V_th × R_L / (R_th + R_L) = 24 × 4 / 10 = 9.6 V. The rest of the source voltage drops across the internal resistance: 2.4 A × 6 Ω = 14.4 V, and 14.4 V + 9.6 V = 24 V, confirming the result.

Key idea: in a Thevenin circuit, the load sits in series with the internal resistance, so the same current flows through both and is set by the total resistance.

The current through the load is I = V_th / (R_th + R_L) = 24 / (6 + 4) = 2.4 A. The voltage across the load is V_L = I × R_L = 2.4 A × 4 Ω = 9.6 V. You can also see this as a voltage divider: V_L = V_th × R_L / (R_th + R_L) = 24 × 4 / 10 = 9.6 V. The rest of the source voltage drops across the internal resistance: 2.4 A × 6 Ω = 14.4 V, and 14.4 V + 9.6 V = 24 V, confirming the result.