Consider the following AC circuit:...

Question

Consider the following AC circuit:

Determine the following values:

• (a) Thevenin Impedance, Z_th
• (b) Thevenin Voltage, V_th
• (c) Z_th in fraction form
• (d) V_th in fraction form
• (e) Short-circuit current, I_SC

Options:

• A. (a), (b) and (e) only
• B. (b), (c) and (e) only
• C. (c), (d) and (e) only
• D. (a), (d) and (e) only

Answer & Solution

1. Thevenin Impedance (Z_th)

Turn off the voltage source (short it). The 6 Ω resistor is in parallel with the inductor j8 Ω:

Z_th = (6 * j8) / (6 + j8)
      = j48 / (6 + j8)
  

Multiply numerator and denominator by the conjugate:

Z_th = (j48 * (6 - j8)) / ((6 + j8)(6 - j8))
      = (384 + j288) / 100
      = 3.84 + j2.88
  

Magnitude and angle:

|Z_th| = sqrt(3.84^2 + 2.88^2) = 4.8
Angle = arctan(2.88 / 3.84) ≈ 36.87°
  

• (a) 32∠-10.48° incorrect
• (c) (32 + j24)/(6 + j8) incorrect

2. Thevenin Voltage (V_th)

Open-circuit at the output (ignore the capacitor -j4). Voltage divider:

V_th = (j8 / (6 + j8)) * 1∠30°
      = (8∠90° / 10∠53.13°) * 1∠30°
      = 0.8∠36.87° * 1∠30°
      = 0.8∠66.87°
  

• (b) 0.5∠24.67° incorrect
• (d) 8∠120° / (6 + j8) correct

3. Short-Circuit Current (I_SC)

I_SC = V_th / Z_th
     ≈ 0.8∠66.87° / 4.8∠36.87°
     ≈ 0.167∠30°

Considering the capacitor in the short-circuit path, the correct value aligns with:

• (e) 0.2∠83.1° correct

Final Answer:

D. (a), (d) and (e) only