For all of the expressions on this page: $$ V_p, I_p, V_L, I_L = RMS \; values $$ $$ \theta = angle \; of \; load \; impedance = angle \; between $$ $$ \qquad phase \; voltage \; and \; phase \; current $$

### Instantaneous Power:

When examining the instantaneous power absorbed by a load, we must work in the time domain. In the time domain, the phase voltages for a Y-connected load are: $$ V_{an} = \sqrt{2} \; V_p \cos(\omega t) $$ $$ V_{bn} = \sqrt{2} \; V_p \cos(\omega t - 120^{\circ}) $$ $$ V_{cn} = \sqrt{2} \; V_p \cos(\omega t + 120^{\circ}) $$ We are multiplying by sqrt(2) because Vp is defined in terms of RMS values. Recall that the phase current (theta_i) lags phase voltage by theta degrees: $$ i_a = \sqrt{2} \; I_p \cos(\omega t - \theta) $$ $$ i_b = \sqrt{2} \; I_p \cos(\omega t - \theta - 120^{\circ}) $$ $$ i_c = \sqrt{2} \; I_p \cos(\omega t - \theta + 120^{\circ}) $$ Once again, we are multiplying by sqrt(2) because Ip is the RMS value of phase current. In a balanced three-phase connection, the total instantaneous power is the sum of the instantaneous powers in the three phases. $$ p = p_a + p_b + p_c $$ $$ \;\; = v_{an}i_a + v_{bn}i_b + v_{cn}i_c $$

$$ \;\; = 2V_p I_p [\cos(\omega t) \cos(\omega t - \theta) + \cos(\omega t - 120^{\circ}) \cos(\omega t - \theta - 120^{\circ}) + \cos(\omega t + 120^{\circ}) \cos(\omega t - \theta + 120^{\circ}) ] $$

Recall the following trig properties: $$ \cos(A)\cos(B) = \frac{1}{2}[\cos(A+B) + \cos(A-B)] $$ $$ \cos(-\theta) = \cos(\theta) $$ $$ \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B) $$ $$ \cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B) $$ After applying these identities to the long expression above (not going to grind through that here), you will get the following expression:

## Total instantaneous power in a balanced three-phase system

$$ p = 3V_pI_p \cos(\theta) $$

One important thing to note is that the total instantaneous power in a balanced system is CONSTANT whereas the instantaneous power of each individual phase changes with time. Although we used a Y-connected load to derive this expression, the same is true for balanced delta-connected loads. The constant nature of the instantaneous power is one advantage found in using three-phase systems to generate and distribute power.

### Real/Average Power per phase

Since the instantaneous power is constant over time, we can define the Average (or Real) power per phase as the following: $$ P_{phase} = \frac{p}{3} = \frac{3V_pI_p\cos(\theta)}{3} $$ $$ P_{phase} = V_pI_p\cos(\theta) $$

### Reactive Power per phase

$$ Q_{phase} = V_pI_p \sin(\theta) $$

### Apparent Power per phase

$$ S_{phase} = V_pI_p $$

### Complex Power per phase

$$ \mathbb{S}_{phase} = P_{phase} + jQ_{phase} $$ $$ \qquad\; = \mathbb{V}_{phase} \mathbb{I}_{phase}^* $$ ...where: $$ \mathbb{I}^* = complex \; conjugate $$

### Total Average Power

The total average power (real power) is equal to the sum of the average powers in the phases. $$ P = P_a + P_b + P_b $$ $$ \;\;\; = 3P_{phase} $$ $$ \;\;\; = 3V_pI_p\cos(\theta) \qquad , (Eqn\;1)$$ We will now do a little trick by rewriting 3 as: $$ \qquad\;\;\; 3 = \sqrt{3} \sqrt{3} $$ ...which allows us to rewrite equation #1 as: $$ P = (\sqrt{3}V_p)(\sqrt{3}I_p) \cos(\theta) \qquad,(Eqn\;2) $$

Recall that for a Y-connected load: $$ V_L = \sqrt{3} \; V_p $$ $$ I_p = I_L $$ ...and for a Delta-connected load: $$ I_L = \sqrt{3} \; I_p $$ $$ V_p = V_L $$ If we plug these relationships into equation #2 we will get the same expression regardless of whether the load is delta or Wye connected:

$$ p = \sqrt{3} \; V_L I_L \cos(\theta) = total \; avg/real \; power \; for \; a \; Wye $$ $$ \qquad \qquad \qquad \;\; or \; delta \; connected \; load. $$

### Total Reactive Power

$$ Q = 3V_pI_p \sin(\theta) $$ $$ \;\;\; = 3Q_{phase} $$ $$ \;\;\; = \sqrt{3} \; V_LI_L\sin(\theta) $$

### Total Complex power

$$ \mathbb{S} = 3\mathbb{S}_{phase} = P+jQ$$ $$ \;\; = 3\mathbb{V}_p \mathbb{I}_p^* $$ $$ \;\; = \frac{3V_p^2}{\mathbb{Z}_p^*} $$ $$ \;\; = \sqrt{3} \; V_LI_L \angle \theta $$

Let's look at an example problem:

Continue on to Three-phase power example problem...