# Balanced Three-phase Power Analysis

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...