Test Index

GATE Electrical Engineering (EE) 2011 Solved Paper

© examsnet.com

Question : 11

Total: 65

V_s = 1∠0° C
I_s = √2 ∠π/4
I_RL = √2 ∠-π/4
The current I_C in the figure above is

-j 2 A
−j‌
1
√2
A
+j‌
1
√2
A
+j 2 A

Go to Question:

Prev Question

Next Question