In Thevenin analysis with V_th = 18 V and R_th = 9 Ω, what is the short-circuit current I_sc?

In Thevenin analysis with V_th = 18 V and R_th = 9 Ω, what is the short-circuit current I_sc?

• A. 3 A
• B. 2 A
• C. 1 A
• D. 0.5 A

Answer: (B)

Short-circuit current is determined by the Thevenin voltage divided by the Thevenin resistance when the output is shorted. With V_th = 18 V and R_th = 9 Ω, the short sits across just the 9 Ω, so I_sc = 18 V / 9 Ω = 2 A. The current flows from the positive terminal through the short toward the negative terminal.

Short-circuit current is determined by the Thevenin voltage divided by the Thevenin resistance when the output is shorted. With V_th = 18 V and R_th = 9 Ω, the short sits across just the 9 Ω, so I_sc = 18 V / 9 Ω = 2 A. The current flows from the positive terminal through the short toward the negative terminal.