The current through a load connected to a Thevenin equivalent with complex impedance Z_th and Z_load is given by which expression?

The current through a load connected to a Thevenin equivalent with complex impedance Z_th and Z_load is given by which expression?

• A. I = V_th / Z_load
• B. I = V_th / (Z_th - Z_load)
• C. I = V_th * (Z_th + Z_load)
• D. I = V_th / (Z_th + Z_load)

Answer: (D)

The question tests how current flows in a Thevenin circuit when a load is connected. In the Thevenin model, you have an ideal voltage source V_th in series with an impedance Z_th. The load Z_load is connected across that output, so all three elements sit in series and share the same current. For AC or complex impedance analysis, Ohm’s law uses impedance in place of resistance, and the total impedance seen by the source is Z_th plus Z_load. Therefore the current is the source voltage divided by the total impedance: I = V_th / (Z_th + Z_load). This makes sense intuitively: if the load becomes larger, the total impedance grows and the current drops; if the load is small, the current increases. The plus sign is essential here because the impedances add in series, not subtract or multiply.

The question tests how current flows in a Thevenin circuit when a load is connected. In the Thevenin model, you have an ideal voltage source V_th in series with an impedance Z_th. The load Z_load is connected across that output, so all three elements sit in series and share the same current. For AC or complex impedance analysis, Ohm’s law uses impedance in place of resistance, and the total impedance seen by the source is Z_th plus Z_load. Therefore the current is the source voltage divided by the total impedance: I = V_th / (Z_th + Z_load).

This makes sense intuitively: if the load becomes larger, the total impedance grows and the current drops; if the load is small, the current increases. The plus sign is essential here because the impedances add in series, not subtract or multiply.