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Q#1
Transient Analysis
GATE EE 2006
MCQ
+2 marks
-0.66 marks
In the circuit shown in the figure, the current source I = 1A, voltage source V = 5V, 
, 
, 
In steady state, the currents (in A) through R3 and the voltage source V respectively will be
1, 4
5, 1
5, 2
5, 4
Explanation:
Since no switching action takes place, so the circuit is in steady state.
In steady state capacitor acts as open circuit while inductor acts as short circuit.
So redrawing the circuit:
On applying KCL at node A,
Applying KVL in 2nd loop
Current in 
Current in voltage source=4A
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