Believe it or not as a kid in grade school I used to hate math, particularly algebra. None of my teachers were able to decipher its complexities and render it comprehensible to me or the majority of my classmates. Then in high school everything changed. I had Mr. Coleman for freshman algebra, and he had a way of making it both understandable and fun, in a challenging kind of way. With 40 years of teaching under his belt, Mr. Coleman knew exactly how to convey the required information in an understandable manner, and to this day I find his insights useful in solving engineering calculations.
Last time we began our discussion on Ohm’s Law and how it may be applied to our example circuit to solve for the electrical current flowing through it. Let’s continue our discussion to see how the Law applies to only one part of the circuit. Then, we’ll use a little algebra to show how the output voltage of an unregulated power supply is affected by changes in **R**_{Total}.
## Figure 1
To help us see things more clearly, in Figure 1 we’ll cover up the inside workings of the unregulated power supply side of the circuit and concentrate on the external supply part of the circuit alone. Since **R**_{Total} is connected to the terminals of the power supply, the voltage applied to **R**_{Total }is the same as the power supply output voltage, **V**_{Output}.
In my previous article, we learned that according to Ohm’s Law, the current flowing through a resistance is equal to the voltage applied to it, divided by the resistance. The fact that **R**_{Total} is connected to the two output terminals like we see in Figure 1, allows us to use Ohm’s law to solve for the electrical current, **I**, flowing through _{ }R_{Total}:
**I = V**_{Output} ÷ R_{Total}
Now let’s pull the cover off of the unregulated power supply again to see what’s going on within the circuit as a whole.
## Figure 2
In Figure 2 we can see that the current, **I**, flowing through **R**_{Total} is the same current flowing through the balance of the circuit. In the preceding blog we found that value to be:
**I = V**_{DC} ÷ (R_{Internal }+ R_{Total})
We can combine the above two equations for **I** to develop an algebraic relationship between **V**_{Output} and **R**_{Internal}, **R**_{Total}, and **V**_{DC}:
**V**_{Output} ÷ R_{Total } = V_{DC} ÷ (R_{Internal }+ R_{Total})
Then, by rearranging terms and applying the cross multiplication principle of algebra we can solve for **V**_{Output}. This involves multiplying both sides of the equation by **R**_{Total:}
**V**_{Output} = R_{Total} × (V_{DC} ÷ (R_{Internal} +R_{Total}))
This equation tells us that although **R**_{Internal} doesn’t fluctuate, **V**_{Output} will fluctuate when **R**_{Total} does. This fact is demonstrated in our equation when we make use of algebra. That is to say, when a term changes on one side of the equation, it causes the other side of the equation to change as well. In this case, when **R**_{Total} changes, it causes **V**_{Output} to change in proportion to the fixed values of **V**_{DC} and **R**_{Internal}.
Next time we’ll look at another shortcoming of unregulated power supplies, more specifically, how one supply can’t power multiple electrical circuits comprised of different voltages.
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