How is short-circuit current calculated using Thevenin equivalents?

How is short-circuit current calculated using Thevenin equivalents?

• A. I_SC = V_th * Z_th
• B. I_SC = V_th / Z_th
• C. I_SC = Z_th / V_th
• D. I_SC = V_th - Z_th

Answer: (B)

When you replace the network with its Thevenin equivalent, you have a voltage source V_th in series with an impedance Z_th. If the terminals are shorted, the only impedance in the path is Z_th, so the short-circuit current follows Ohm’s law: I_SC = V_th / Z_th. In AC, treat V_th and Z_th as phasors, so I_SC = V_th / Z_th, with magnitude |I_SC| = |V_th| / |Z_th| and angle equal to arg(V_th) − arg(Z_th). This is the correct relation because the current through a series path is the source voltage divided by the series impedance; other forms that multiply, divide in the opposite order, or subtract do not represent the short-circuit current.

When you replace the network with its Thevenin equivalent, you have a voltage source V_th in series with an impedance Z_th. If the terminals are shorted, the only impedance in the path is Z_th, so the short-circuit current follows Ohm’s law: I_SC = V_th / Z_th. In AC, treat V_th and Z_th as phasors, so I_SC = V_th / Z_th, with magnitude |I_SC| = |V_th| / |Z_th| and angle equal to arg(V_th) − arg(Z_th). This is the correct relation because the current through a series path is the source voltage divided by the series impedance; other forms that multiply, divide in the opposite order, or subtract do not represent the short-circuit current.