Why divide voltage by impedance only for the resistor in an RL circuit

[teng125]

v(rt) = 125mV cos (wt + 60)
R=250 ohms
L = 250mH
find i(t)
in the circuit there is a inductor in series wif the resistor

the answers i(t)=0.5 cos (wt+60)

may i know why is it the inductor not using to form z and then only divide by v(rt) ??why only the resistor??

 

[Astronuc]

The resistor impedance is 'real', i.e. voltage and current are in phase. The inductor and capacitor each have reactance or 'complex' impedance. i.e voltage and current are out of phase.

 

[teng125]

so when only the inductor will be included??

 

[Astronuc]

Edit: see my subsequent post.

The phase angle of the inductor should be part of the answer since it does have complex impedance. Is the answer from the textbook?

One uses I = V/Z = VZ*/ZZ* and this can be written in phasor form.

http://en.wikipedia.org/wiki/Electrical_impedance#Magnitude_and_phase_of_impedance

Z = R + jwL
http://en.wikipedia.org/wiki/Electrical_impedance#In_series

 

[teng125]

yes,it is from the book

 

[teng125]

for z = 250 + j250 since w=1000 right??
then v(rt) = 125mV cos (wt + 60)

so can i divide them (v/z) ??

 

[Astronuc]

Ah, I think I now understand what's going on here! I apologize for the confusion.

Looking at the voltage v(rt) = 125mV cos (wt + 60)

I think that should be v_R(t) = 125mV cos (wt + 60), which is the voltage across the resistor.

Then the current must be i(t) = v_R(t)/R, and because the inductor and resistor are in series, the same current MUST pass through both.

Now the voltage across the resistor AND inductor is

v(t) = v_R(t) + v_L(t) = i(t) R + i(t) * jwL.

 

[teng125]

oo...okok thanx