- #1
mmmboh
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Hi here is the problem:
What I did was Z=[(XL-Xc)2+R2]1/2 = 5 ohm
and then V/R = I = 110(2)1/2/5=31.1A
For part B) I did V=L(di/dt)=-Lwsin(wt), and the peak amplitude is when sin(wt)=-1 and is 377V.
Can someone tell me if what I did is right, or what is wrong?
Thanks.
and then V/R = I = 110(2)1/2/5=31.1A
For part B) I did V=L(di/dt)=-Lwsin(wt), and the peak amplitude is when sin(wt)=-1 and is 377V.
Can someone tell me if what I did is right, or what is wrong?
Thanks.
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