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The question requests that I solve the circuit below for v0(t). I'm solving for the voltage over the inductor. I'm getting a result that's close to what I expect, however I think the phase angle of the voltage is slightly off (some friends of mine said they all got 46°, whereas I am consistantly getting 45.02°).
My equations:
ix=(16∠45°)/(R)=(16∠45°)/(4E3) = 4E-3∠45°
KVL of second loop:
0=0.754j(i1+8E-3∠45°)+4E3(4E-3∠45°)-53.05j(i1)
i1=0.306∠-44.98°
v0=0.754j(0.306∠-44.98°) = 0.231∠45.02°
v0(t)= 0.231cos(377t+45.02°)V
Image of the circuit:
My equations:
ix=(16∠45°)/(R)=(16∠45°)/(4E3) = 4E-3∠45°
KVL of second loop:
0=0.754j(i1+8E-3∠45°)+4E3(4E-3∠45°)-53.05j(i1)
i1=0.306∠-44.98°
v0=0.754j(0.306∠-44.98°) = 0.231∠45.02°
v0(t)= 0.231cos(377t+45.02°)V
Image of the circuit: