#12 Derive line-current subtraction and calculate balanced RMS power

Below are all the lines written in the notebook together with the full narration transcript.

1. 1. Recall the balanced wye relations

   

   Recall a balanced ABC wye set, with RMS phase magnitudes:

   $\displaystyle V_{L}=\sqrt{3} V_{p}$

   Now use a delta (Δ) load with the same positive phase sequence.

   Three identical impedances connect a to b, b to c, and c to a.

   The delta load has no shared neutral node; this ideal circuit description is not wiring advice.

   The square-root factor now relates balanced current magnitudes:

   $\displaystyle I_{L}=\sqrt{3} I_{p}$

   Narration transcript

   In the previous lesson we connected three balanced phasors to a Y-Y circuit and discovered that the line voltage equals square root of three times the phase voltage. Now we flip the topology to delta. Three impedances form a closed triangle. There is no neutral. The square root of three factor reappears, but this time on the current side.

2. 2. Define delta terminals and current directions

   

   Use the reviewed branch-direction card; the original circuit picture has outward line arrows inconsistent with the inward reference used here.

   Three equal impedances form the closed delta loop:

   $\displaystyle Z_{\mathrm{ab}}=Z_{\mathrm{bc}}=Z_{\mathrm{ca}}=Z_{\Delta }$

   There is no shared central node in this delta load.

   Define each line current into its load terminal; branch currents use the cyclic directions shown in the card.

   Each branch is connected across two lines, so its RMS voltage magnitude is:

   $\displaystyle V_{p}=V_{L}$

   The branch from a to b uses the same ordered line-voltage reference:

   $\displaystyle V_{\mathrm{ab}}=V_{a}-V_{b}$

   Narration transcript

   Here is the standard delta connection. Three impedances Z delta sit between the three line terminals a, b, and c, forming a closed triangle. Notice what is missing -- there is no neutral point and no fourth wire. Each impedance carries a phase current, and each line carries a line current. The fundamental difference from Y is that in delta, the phase voltage equals the line voltage. The same V a b that drives the line in Y now drives one entire impedance arm of the delta.

3. 3. Fix a positive-sequence branch-current reference

   

   Branch references are I_ab from a to b, I_bc from b to c, and I_ca from c to a.

   Assume balanced sinusoidal terminal voltages and three identical impedances. Then branch currents have equal RMS magnitude I_p and positive ABC spacing.

   Narration transcript

   In delta, the three phase currents flow inside the triangle: I a b flows from a to b through the first impedance, I b c from b to c, and I c a from c to a. For a balanced positive sequence load, these three currents form their own balanced set, equal in magnitude and spaced one hundred twenty degrees apart -- just like the voltages were in the previous lesson.

4. 4. Apply KCL with line current into the load

   

   Define I_a as entering the delta load at terminal a.

   At a, I_a and I_ca enter while I_ab leaves:

   $\displaystyle I_{a}+I_{\mathrm{ca}}=I_{\mathrm{ab}}$

   Rearrange KCL without reversing a current reference:

   $\displaystyle I_{a}=I_{\mathrm{ab}}-I_{\mathrm{ca}}$

   The reference image is the KCL card, not a new phasor diagram; derive the subtraction in the next lines.

   Narration transcript

   The line current I a is the current actually flowing through the wire connected to terminal a. By Kirchhoff's current law applied at node a, the current flowing in equals the currents leaving -- so I a equals I a b minus I c a. Two phase currents subtract to give one line current. The same geometric question returns -- what does this subtraction look like on the phasor plane?

5. 5. Subtract phase-current phasors

   

   Use fixed complex RMS phasors, a positive sequence and inward line-current references.

   Choose the branch current as the zero-angle reference:

   $\displaystyle I_{\mathrm{ab}}=I_{p}∠0^{\circ}$

   The c-to-a branch leads this reference by 120°:

   $\displaystyle I_{\mathrm{ca}}=I_{p}∠120^{\circ}$

   Negating a phasor rotates it by 180°:

   $\displaystyle -I_{\mathrm{ca}}=I_{p}∠\left(-60^{\circ}\right)$

   The two equal vectors have a 60° common-origin separation; the corresponding head-to-tail triangle has a 120° interior angle.

   Subtract in rectangular coordinates:

   $\displaystyle I_{a}=I_{p}\left(\frac{3}{2}-\frac{j \sqrt{3}}{2}\right)$

   Compute the magnitude from perpendicular components:

   $\displaystyle |I_{a}|=I_{p} \sqrt{\left(\frac{3}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}=\sqrt{3} I_{p}$

   The real part is positive and imaginary part negative, so the result lies in quadrant IV:

   $\displaystyle ∠I_{a}=-30^{\circ}$

   For these stated references:

   $\displaystyle I_{a}=\left(\sqrt{3} I_{p}\right)∠\left(-30^{\circ}\right)$

   Narration transcript

   Watch the construction. I a b sits along the reference axis. I c a sits at plus one hundred twenty degrees. To form I a equals I a b minus I c a, we flip I c a to its negative, swinging it down to minus sixty degrees, and add it tip-to-tail to I a b. The two equal-length vectors form a rhombus. The diagonal from the origin to the far corner is exactly I a. With both sides of length I phase and an internal angle of sixty degrees, the cosine rule gives a diagonal length of square root of three times I phase. The diagonal lands at minus thirty degrees. So I a lags I a b by thirty degrees, with magnitude root three I phase.

6. 6. Separate the current magnitude and angle

   

   Separate the RMS magnitude relation from its phase-angle condition.

   For a balanced delta load:

   $\displaystyle I_{L}=\sqrt{3} I_{p}$

   With I_ab as reference, the three line-current angles are:

   $\displaystyle ∠I_{a}=-30^{\circ}, ∠I_{b}=-150^{\circ}, ∠I_{c}=90^{\circ}$

   The 30° lag is relative to the corresponding branch current, not directly to its branch voltage.

   In positive sequence, wye gives a 30° line-voltage lead; delta gives a 30° line-current lag under the chosen references.

   Narration transcript

   Here is the symmetric headline. In delta, the line current magnitude is square root of three times the phase current magnitude, but it lags by thirty degrees instead of leading. The other two line currents I b and I c follow the same pattern, each rotated by another one hundred twenty degrees. Three line currents form a second balanced star -- same root three magnitude, rotated minus thirty degrees from the phase star. This is the mirror image of the Y-Y voltage result.

7. 7. Relate branch voltage to line voltage

   

   Now identify the voltage across each delta branch.

   Each impedance arm is connected directly between two line terminals.

   The branch and line-to-line voltages use the same ordered pair of terminals.

   Delta branch and line RMS magnitudes are equal:

   $\displaystyle V_{p}=V_{L}$

   Do not multiply delta line voltage by a square-root factor to obtain its branch voltage.

   In balanced systems, wye introduces the square root of three in voltage magnitudes and delta in current magnitudes.

   Narration transcript

   Now look at the voltage side in delta. Each impedance arm sits directly between two line terminals. So the voltage across one phase impedance is exactly a line-to-line voltage. In delta, V phase equals V line. There is no square root of three on the voltage side. Delta and Y are duals: Y puts root three on voltages, delta puts root three on currents.

8. 8. Use balanced sinusoidal RMS power

   

   Assume balanced sinusoidal voltages and currents; all voltage and current magnitudes below are RMS.

   Average real power in line quantities:

   $\displaystyle P_{\mathrm{total}}=\sqrt{3} V_{L} I_{L} \cos \left(\theta \right)$

   The same result follows from summing the three equal phase powers:

   $\displaystyle P_{\mathrm{total}}=3 V_{p} I_{p} \cos \left(\theta \right)$

   For delta substitute the current relation:

   $\displaystyle I_{p}=\frac{I_{L}}{\sqrt{3}}$

   θ is the load-impedance angle between a branch voltage and its own current, not directly between V_ab and I_a. The formula is not universal for arbitrary unbalanced or nonsinusoidal loads.

   Narration transcript

   Now the most useful result of all. Total three-phase average power can be written in line quantities as square root of three times V line times I line times cosine theta, where theta is the impedance angle. This formula works for both Y and delta connections -- exactly the same. Although V phase and I phase look different in Y versus delta, the line quantities V L and I L combine in such a way that the total power formula is universal. It does not depend on which connection you choose.

9. 9. Calculate the 208 V delta load

   

   Use the original balanced delta example; keep exact values until the final rounding.

   The line-voltage RMS magnitude is:

   $\displaystyle V_{L}=208 V$

   Each inductive branch impedance is:

   $\displaystyle Z_{\Delta }=\left(30+j 15\right) \Omega$

   Its magnitude is:

   $\displaystyle |Z_{\Delta }|=\sqrt{30^{2}+15^{2}}=\sqrt{1125}\approx 33.54 \Omega$

   The impedance angle is:

   $\displaystyle \theta =\arctan \left(\frac{15}{30}\right)\approx 26.57^{\circ}$

   In delta the branch has the full line voltage:

   $\displaystyle V_{p}=V_{L}=208 V$

   Branch current lags its own branch voltage by θ; its RMS magnitude is:

   $\displaystyle I_{p}=\frac{208}{\sqrt{1125}}\approx 6.20 A$

   The line-current RMS magnitude is:

   $\displaystyle I_{L}=\sqrt{3} I_{p}\approx 10.74 A$

   The inductive load has lagging power factor:

   $\displaystyle \cos \left(\theta \right)\approx 0.894$

   Using the rounded source factors gives about 3459 W; keeping exact values gives 3461.12 W. Both round to:

   $\displaystyle P_{\mathrm{total}}\approx 3.46 \left[\mathrm{kW}\right]$

   The earlier wye example used 120 V per phase, so its line voltage was about 207.85 V, rounded to 208 V here; equality of power is approximate in that comparison.

   Exact balanced terminal equivalence at the same line voltage requires:

   $\displaystyle Z_{\Delta }=3 Z_{Y}$

   Narration transcript

   Let us put numbers on it. Take a delta-connected load with line voltage two hundred and eight volts r m s. Each leg of the delta is Z delta equals thirty plus j fifteen ohms. First, the magnitude and angle of Z delta: the magnitude is square root of one thousand one hundred twenty five, which is thirty three point five four ohms. The angle is arctangent of fifteen over thirty, which is twenty six point five seven degrees. Phase current: in delta, V phase equals V line equals two hundred and eight volts. So I phase equals two hundred and eight divided by thirty three point five four, giving six point two zero amperes r m s, lagging by twenty six point five seven degrees. Line current: in delta, I line equals root three times I phase, which is one point seven three two times six point two zero, giving ten point seven four amperes r m s. Cosine of twenty six point five seven degrees is zero point eight nine four. Total power: root three times two hundred and eight times ten point seven four times zero point eight nine four equals about three thousand four hundred fifty eight watts, or three point four six kilowatts. Compare with the Y-Y example from the previous lesson -- same line voltage, same total power. Y or delta, the externally measurable behavior is identical when the loads are equivalent.

10. 10. Interpret the ideal rotating-field model

    

    Now interpret the rotating-field idea as an ideal model, not a literal engineering winding design.

    Use equal sinusoidal currents separated by 120° in time and equal effective coil axes separated by 120° in space.

    In the ideal linear model their field vectors sum to a constant-magnitude vector rotating at constant angular speed.

    The field can rotate with stationary windings; this does not assert that a generator supplies electrical power without mechanical input.

    The reference still is the summary card, not the source 3D coil geometry. Field speed and mechanical rotor speed are not generally identical; single-phase motor starting also needs separate treatment.

    Narration transcript

    Here is why three-phase systems matter so much. Three currents, each shifted by one hundred twenty degrees in time, flow through three coils, each placed one hundred twenty degrees apart in space. Their magnetic vectors add up to a single vector that rotates in the stator at constant speed. No mechanical motion, only electrical phase shifts, produce a smoothly rotating magnetic field. This is the heart of every induction motor and every three-phase generator.

11. 11. Keep topology and balance conditions explicit

    

    Keep the balanced delta result and its reference directions together.

    RMS relations for delta:

    $\displaystyle I_{L}=\sqrt{3} I_{p}, V_{L}=V_{p}$

    For balanced sinusoidal Y or Δ loads:

    $\displaystyle P_{\mathrm{total}}=\sqrt{3} V_{L} I_{L} \cos \left(\theta \right)$

    Rotating-field intuition additionally needs appropriate spatial windings; electrical phase spacing alone is not a machine model.

    Next: the Laplace transform definition, unit step, impulse and basic transform pairs.

    Narration transcript

    Three takeaways for delta. Line current is root three times phase current, lagging by thirty degrees, while line voltage equals phase voltage. Total power is root three V line I line cosine theta -- the same universal formula whether the load is Y or delta. And the same three-phase symmetry that gave us clean equations also gives us rotating magnetic fields in machines. Next we leave AC steady state and start the Laplace transform.

Source video: Circuit Theory-2 #12 | Delta Connections and Three-Phase Power (7:55)