In the given circuit with the shown ideal 5V DC source, the magnitude of the total current drawn from the source at a steady-state is
Right Answer is:
10 A
SOLUTION
Under steady-state conditions, the inductor is replaced by short-circuited. Therefore, a 1 Ω resistor that is connected parallel to the inductor is neglected.
Hence total resistance of the circuit will be
Rtotal = 1Ω || 1Ω = (1 × 1)/(1 + 1)
Rtotal = 0.5
Therefore current drawn by the 5 V source is
I = V/R = 5/0.5
I = 10 A