# 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

#### 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

R_{total} = 1Ω || 1Ω = (1 × 1)/(1 + 1)

R_{total} = 0.5

Therefore current drawn by the 5 V source is

I = V/R = 5/0.5

**I = 10 A**