We'll now start working through some problems where we use our knowledge of electric power flow and apply it to simple networks. We will walk through some theory and simple examples first, then at the bottom of the page are a couple of videos that you can watch to see additional examples. The next page of the lesson also has a self-evaluation quiz so you can go through as many examples as you would like to get practice.

In this section we will start by calculating power flows for a two-node network with parallel lines, like the one shown below.

In this network, R1 and R2 are the resistances of the transmission lines connecting G and L.

Let's suppose that *W* units of power are injected into the transmission network by the generator at node G. If we ignore transmission losses, we know that two things will happen:

- The customer at node L will consume exactly
*W*units of power, because supply and demand have to balance at all times. - The power will divide between the parallel lines based on the ratio of the (inverse) resistance of each line to the equivalent parallel resistance.

Flow on each of the two transmission lines can be written using the following equations:

Flow on Line 1:

$${\text{F}}_{\text{1}}\text{=W\xd7}\left(\frac{\frac{\text{1}}{\text{R1}}}{\frac{\text{1}}{\text{R1}}\text{+}\frac{\text{1}}{\text{R2}}}\right)$$

Flow on Line 2:

$${\text{F}}_{\text{2}}\text{=W\xd7}\left(\frac{\frac{\text{1}}{\text{R2}}}{\frac{\text{1}}{\text{R1}}\text{+}\frac{\text{1}}{\text{R2}}}\right)$$

We get these equations from what we know about equivalent resistances along parallel lines. The terms in parentheses are called "distribution factors," and those describe the proportion of the total power injected at node G that will flow along each of the two lines.

As an example with actual numbers, let's say that R1 = 0.01 Ω and R2 = 0.05 Ω. 100 MW is injected at node G and consumed at node L. Flows on the two lines would be:

Flow on Line 1:

$${\text{F}}_{\text{1}}\text{=100\xd7}\left(\frac{\frac{\text{1}}{\text{0}\text{.01}}}{\frac{\text{1}}{\text{0}\text{.01}}\text{+}\frac{\text{1}}{\text{0}\text{.05}}}\right)\text{}\approx \text{83}$$

Flow on Line 2:

$${\text{F}}_{\text{2}}\text{=100\xd7}\left(\frac{\frac{\text{1}}{\text{0}\text{.05}}}{\frac{\text{1}}{\text{0}\text{.01}}\text{+}\frac{\text{1}}{\text{0}\text{.05}}}\right)\text{}\approx \text{17}$$

Note that F_{1} + F_{2} = 100 MW, the total amount of power injected into the network. If you add up the flows on the lines connected to the load node in our toy power networks, you will always get the total amount of power injected into the network (and consumed at the load). Note that if you add up the two distribution factors, you will get 1. This is a good way to check your work on two-node networks. If the distribution factors do not add up to 1, and if the flows on the two lines don't add up to the total amount of power injected into the network at node G, then you have done something wrong in your calculations.

Here is an intuitive example that you can use to test yourself. If R1=R2, then (as you might expect) the flow on each line will be identical. If *W* = 100 MW, and if R1=R2=0.05 Ω, use the formulas for the two-node network to convince yourself that this is actually the case. If you don't get flow of 50 MW on each line, then you have made a mistake!

An important thing to realize is that only *relative* resistance matters. The actual resistances, in Ohms, do not matter when calculating the distribution factors or the power flows. Let's suppose that R2 = αR1 in the two-node network. We can then write our power flow equations as:

Flow on Line 1:

$$\begin{array}{l}{\text{F}}_{\text{1}}\text{=W\xd7}\left(\frac{\frac{\text{1}}{\text{R1}}}{\frac{\text{1}}{\text{R1}}\text{+}\frac{\text{1}}{\alpha \text{R1}}}\right)\text{}\\ \text{=W}\times \text{}\left(\frac{1}{1+\frac{1}{\alpha}}\right)\end{array}$$

Flow on Line 2:

$$\begin{array}{l}{\text{F}}_{\text{2}}\text{=W\xd7}\left(\frac{\frac{\text{1}}{\alpha \text{R1}}}{\frac{\text{1}}{\text{R1}}\text{+}\frac{\text{1}}{\text{\alpha R1}}}\right)\text{}\\ \text{=W\xd7}\left(\frac{\frac{\text{1}}{\alpha}}{\text{1+}\frac{\text{1}}{\text{\alpha}}}\right)\end{array}$$

Note that both R1 and R2 have totally disappeared, and all we are left with is α, which describes the size of R1 relative to R2.