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**1. Homework Statement**

Use superposition to find [tex]I_L[/tex]

**2. Homework Equations**

-The overall current is the sum of the currents supplied by each source individually

**3. The Attempt at a Solution**

for source B only (current source is open):

---ohm's law: [tex]\frac{100V}{15k + j25k} = 3.43mA( -59.04 degrees)[/tex]

Here is where I'm having to defend my answer:

for souce A only (voltage source shorted):

---since there is no phase shift labeled, it should be assumed that the current source can be treated as a DC source, treating the inductor as a short, making [tex]I_L[/tex] -20mA

Is this wrong? Should I treat a current source as AC if it has other AC sources involved in the circuit? If this is so I'd just do a current divider to get [tex]I_L[/tex]