# Instantaneous Current in single phase circuit

• Engineering

## 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!

NascentOxygen
Staff Emeritus
Hi Robjob. 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

To put 94.94<-31.26 A into instantaneous form you would do:

I(t) = 94.94cos(314t-31.26) Amps

rude man
Homework Helper
Gold Member
(2.) Figure out the value of the inductor:
jωL = j(0.7) => j(50)L
Recompute ω!