What is the Thevenin equivalent voltage seen by RL in GATE EC 2013 Set-2 Q39?

VTh = 800∠90° V. With RL open, I2=0. The dependent source j40I2=0. Left mesh: I1 = VS/(3+j4) = 100∠53.13°/5∠53.13° = 20∠0° A. VL1 = j4×I1 = j80 V. Since no current flows through j6Ω and 5Ω, VTh = 10VL1 = 10×j80 = j800 = 800∠90° V.

How does the dependent current source j40I2 affect the Thevenin voltage in GATE EC 2013 Set-2 Q39?

When RL is open-circuited to find VTh, the right mesh current I2 = 0. Since the dependent current source value is j40I2, it becomes j40×0 = 0, acting as an open circuit. This simplifies the left mesh so that VS drives only the 3Ω and j4Ω series impedance, making I1 = VS/(3+j4) = 20∠0° A.

Why is VTh = 800∠90° and not 100∠90° in GATE EC 2013 Set-2 Q39?

VL1 = j4×I1 = j4×20 = j80 V. The dependent voltage source in the right mesh is 10VL1 = 10×j80 = j800 = 800∠90° V. Since no current flows through j6Ω and 5Ω (I2=0), the full 800∠90° appears across the open RL terminals as VTh. The factor of 10 in the dependent source amplifies VL1, giving magnitude 800, not 100.

Thevenin Voltage Dependent Source VS=100∠53.13° | GATE EC 2013 Set-2 Q39

EC > Gate 2013 SET-2 > Network Theorems

In the circuit shown below, if the source voltage V_S = 100∠53.13° V then the Thevenin's equivalent voltage in Volts as seen by the load resistance R_L is

100∠90°

800∠0°

800∠90°

100∠60°

Correct : C

With R_L open, I₂ = 0. The dependent current source j40I₂ = 0, acting as an open circuit. The left mesh reduces to:
V_S = (3 + j4)I₁
I₁ = 100∠53.13° / 5∠53.13° = 20∠0° A
V_L1 = j4 × I₁ = j4 × 20 = j80 V = 80∠90° V
With no current through j6Ω and 5Ω (since I₂ = 0), the open-circuit voltage across R_L equals the dependent voltage source:
V_Th = 10V_L1 = 10 × j80 = j800 V = 800∠90° V
Correct answer: C — 800∠90°.

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Thevenin voltage AC circuit GATE 2013 GATE EC 2013 Set-2 Q39 VS=100 angle 53.13 Thevenin equivalent GATE network theorems Thevenin phasor GATE EC Thevenin voltage 100 angle 90 degrees GATE AC circuit phasor analysis GATE EC 2013

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