Ohms law

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gar

Senior Member
Location
Ann Arbor, Michigan
Occupation
EE
170711-0920 EDT

Sahib:

You need to get down to basics. Somehow you are not corrrectly looking at a basic loop equation that sums to 0.

I set up a simple experiment with an AC voltage source, 60 Hz power line at reduced voltage, a 75 ohm resistor in series with a 175 VA 120 V transformer primary, and an oscilloscope. The reason for the reduced voltage was to minimize transformer saturation current. The voltage channel used a 10x probe. Thus, peak voltage was about 18 V. Peak current about 5 mA. As a side note: the distortion of the voltage peak is not seen in the current waveform because of the filter action of the inductance.

In the scope plot you see a single shot display of a steady state condition of the series RL and AC source circuit. Note the lagging phase shift of the current.

mivey has told you and you should have on your own known that there is voltage drop across the resistance when the AC source voltage is zero. That this is true is easily seen in the scope plot. Following is the plot with blue being source voltage, and red being being the voltage across the 75 ohm resistor. There is also about 1.5 ohms resistance internal to the transformer. The phase shift is about 45 deg.

DS2_QuickPrint77M.JPG




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mivey

Senior Member
mivey: After some deep meditation on your words about voltage of resistor is exactly opposite to that of inductor when source voltage is zero in RL circuit, I found that it contradicts KVL when values of inductive reactance and resistance are not same.
Stop meditating and start calculating. Let me help you get started with a circuit and the results from three different times. Ignoring rounding errors and hopefully I have not fat-fingered some calcs:

v_source = 120 Vrms
R = 1
L = 0.00153146709058356
freq = 60
theta = -atan(2*pi()*60 * 0.00153146709058356 / 1) = -30deg = -0.52359819219515 rad

v(t) = sqrt(2)*120 * sin( 2*pi()*60 * t )
i(t) = sqrt(2)*120 / sqrt(1^2+(2*pi()*60 * 0.00153146709058356)^2) * sin( 2*pi()*60 * t + -0.52359819219515 )

v_R(t) = i(t) * 1
v_L(t) = v(t) - v_R(t)


at t = 0 (0 degrees):
v(t) = 0
i(t) = i_R(t) = i_L(t) = -73.4846427800464
p(t) = 0
v_R(t) = -73.4846427800464
p_R(t) = v_R(t) * i_R(t) = 5399.99272451102
v_L(t) = 73.4846427800464
p_L(t) = v_L(t) * i_L(t) = -5399.992725


at t = 0.000138888888888889 (3 degrees):
v(t) = 8.881706294
i(t) = i_R(t) = i_L(t) = -66.72265042
p(t) = -592.61098422899
v_R(t) = -592.61098422899
p_R(t) = v_R(t) * i_R(t) = 4451.91207946539
v_L(t) = 75.6043567171908
p_L(t) = v_L(t) * i_L(t) = -5044.52306369438


at t = 0.00138888888888889 (30 degrees):
v(t) = 84.8528137423857
i(t) = i_R(t) = i_L(t) = 0.0000857424306778816
p(t) = 0.00727548650012971
v_R(t) = 0.0000857424306778816
p_R(t) = v_R(t) * i_R(t) = 7.35176441855134E-09
v_L(t) = 84.852727999955
p_L(t) = v_L(t) * i_L(t) = 0.00727547914836529
 

Besoeker

Senior Member
Location
UK
Stop arguing
This is how is for a restive load,
Twice the frequency and all positive..
I am seriously sight imoaired at the moment.
This is what I meant to post.

1V1A1W_zps788ce72e.jpg.html
 
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