Instantaneous Current in single phase circuit

  • #1
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Homework Statement



SOURCE:
v(t) = (311)cos(314t) [v]

IN SERIES WITH:
R = (0.3) [Ω]
L = j(0.7) [Ω]
"load" = 2.5 + j(1.0) [Ω]


Homework Equations


Find the instantaneous current, and the phasor current.

The Attempt at a Solution


(1.) I first found the frequency of the source:
(314) / (2*pi) ≈ 50 Hz.

(2.) Figure out the value of the inductor:
jωL = j(0.7) => j(50)L = j(0.7) => L = (0.7/50) = 14 [mH]

(3.) I started doing KVL around the loop:
(-311)cos(314t) + (0.3)(i(t)) + (14e-3) (di/dt) ...

Then I didn't know what to do for the "load." It isn't specified if it's a capacitor/inductor/mix, and I'm not sure how to go from 2.5 + j(1.0) [Ω] to something I can use in the time domain? Is it just the real part of that value?
So... (-311)cos(314t) + (0.3)(i(t)) + (14e-3) (di/dt) + 2.5 = 0 ?

---------------------------------------------------------

(1.) For the phasor I found the total impedance:
0.3 + j(0.7) + 2.5 + j(1.0) = 2.8 + j1.7 = 3.27∠31.26 [Ω]

(2.) Then did:
I = V/Z = (311∠0) / (3.27∠31.26) = 94.94∠-31.26 [A]

Does the phasor look correct, and can I back into the instantaneous from the phasor?

Thanks for any help! I'm just getting back into this and I'm pretty rusty!
 

Answers and Replies

  • #2
NascentOxygen
Staff Emeritus
Science Advisor
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Hi Robjob. :welcome:

You haven't provided the question, so it's difficult to say whether your answer is correct. Please include the problem statement.

One error: XL = jwL where w is 314 not 50
 
  • #3
267
10
To put 94.94<-31.26 A into instantaneous form you would do:

I(t) = 94.94cos(314t-31.26) Amps
 
  • #4
rude man
Homework Helper
Insights Author
Gold Member
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(2.) Figure out the value of the inductor:
jωL = j(0.7) => j(50)L
Recompute ω!
 

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