A Thevenin equivalent is V_th = 24 V and R_th = 6 Ω. The load resistor R_L = 4 Ω is connected across the Thevenin source. What is the current through R_L and the voltage across it?

A Thevenin equivalent is V_th = 24 V and R_th = 6 Ω. The load resistor R_L = 4 Ω is connected across the Thevenin source. What is the current through R_L and the voltage across it?

• A. I = 2.4 A; V_L = 9.6 V
• B. I = 3.0 A; V_L = 12.0 V
• C. I = 2.0 A; V_L = 8.0 V
• D. I = 1.8 A; V_L = 7.2 V

Answer: (A)

When a Thevenin source is connected to a load, they form a single loop where the load and the Thevenin resistance share the same current. The current is found from the total resistance in series: I = V_th / (R_th + R_L). Here that’s 24 V / (6 Ω + 4 Ω) = 24 / 10 = 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 via a voltage divider: V_L = V_th × R_L/(R_th + R_L) = 24 × 4/10 = 9.6 V. The remaining voltage drops across the Thevenin resistance (14.4 V), since 2.4 A × 6 Ω = 14.4 V. So the current is 2.4 A and the voltage across the load is 9.6 V.

When a Thevenin source is connected to a load, they form a single loop where the load and the Thevenin resistance share the same current. The current is found from the total resistance in series: I = V_th / (R_th + R_L). Here that’s 24 V / (6 Ω + 4 Ω) = 24 / 10 = 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 via a voltage divider: V_L = V_th × R_L/(R_th + R_L) = 24 × 4/10 = 9.6 V. The remaining voltage drops across the Thevenin resistance (14.4 V), since 2.4 A × 6 Ω = 14.4 V. So the current is 2.4 A and the voltage across the load is 9.6 V.