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DC Electricity – Series Sources and Parallel loads.

Picture 1 – Series Sources

This is the 2nd of 3 parts of DC electrical theory. This is also the hardest of the 3 parts because I am working hard to get us past this and into the “real” stuff.  I am combining several new things in each diagram.

In the last post we ended talking about a case where our load was resistors in series.  In Picture 1, I am still using 2 resistors in series, but this time I am using two different values of resistance to drive the point home.  In the diagram, I am showing two Voltage sources in series.  This is exactly the same thing done in many battery powered devices.  Often in flash lights and many similar devices the batteries are placed into the device with the positive of one battery connected to the negative of the next one.   For example, many older flash lights required 2 D cells.  Each cell puts out 1.5V making a total of 3 Volts.  The lamp in that flash light was a 3 Volt lamp.

In our example I chose one source Voltage of 2.5 Volts and the other Voltage of 4 Volts making the total source Voltage of 6.5 Volts.  Again, I drew this with dimension lines.  This is useful to get a picture in your mind of the voltages adding.  Our total resistance is equal to 10 Ω + 5 Ω or 15 Ω as explained in the last post. Using Ohm’s law, we determine the current flowing is V/R = I and this is  6.5 V/15 Ω = 0.4333 A.   Because the two series resistors were not the same value the Voltage drop across each was also not equal.  The Voltage drop was determined by V = I X R.  This gave the values shown in the diagram.

Picture 2: Series Opposing Sources

In the next picture, I diagram exactly the same circuit, except I reversed the direction of the 2.5 V source.  (The term most often used is I reversed the polarity of the 2.5 V source.)  The math is exactly the same as before except we will be adding a -2.5 V.  Total source voltage is 4 V + -2.5 V = 1.5 V.  The rest of the calculations are exactly the same as the previous example, but return different values because the total source Voltage is different.

Picture 3 – Parallel Resistors

Now that we have series resistors and sources understood, it is time to start complicating things more.  In Picture 3, we are again back to one source but the load consists of two resistors wired in parallel.  The current coming from the source will get divided with part of it going through one resistor and part going through the 2nd resistor.  The old statement about electricity taking the path of least resistance is not true.  Most of it will take the path of least resistance, but some will take the path of the greater resistance. The diagram should make it clear why I insist on drawing the current as arrows running along the conductor path.

This brings us to Kirchhoff’s current law.  The law is simple:  the current going into a node, (connection) must be equal to the current going out of that node.  If you think about it, this is exactly the same rule that my doctor keeps harping about to me.. my caloric intake needs to be equal to my caloric use (out go).  The only thing is the node in the diagram does not have the ability to swell up like my waist does.   Back to a serious note:  A node does not have to be a single connection point an may be a long conductor with many taps on it.

Picture 4: Calculation of Parallel Resistors

Since the source voltage must be dropped by each resistor one way of determining the total current is to calculate the current through each resistor and then add the currents to calculate the total.  However, if we also had another resistor in series with the parallel combination this would not be possible.  (We will run into that situation in real circuits sometimes.)  The way of calculating the total resistance of resistors in parallel is to use the formula shown in Picture 4.   This rule is called the reciprocal of the sum of the reciprocals. Once we know the equivalent resistance, the rest of the calculations is our old friend, Ohm’s law.

Picture 5: Voltage Reference Point.

Our final example is one with a reference point that splits the power sources.  There are two new symbols being introduced in this diagram.  First there is the connection point drawn buy the dark circles.  This just makes it clear that several wires are connected together.   The second new symbol is the triangle.  This just shows that this is the “common” point or reference point of 0 Volts.   This means that the Voltage at point B will be called +15 V and the Voltage at point C will be called -15 Volts.  Also, because this provides a common connection to the whole circuit, the circuit can be calculated in this case as two separate circuits except in the case of the current flowing in the wire between Pt A and the junction of the two resistors.

There are two ways to understand what is happening.  One way is to calculate each 1/2 of the circuit separately.  The bottom circuit would flow 100 mA (0.1 A or 100 one-thousandths of an Ampere).  This current would be flowing from left to right in the wire from pt A to the resistor junction.  The current due to the top circuit would be 50mA flowing from right to left in the wire.   This flow would be counter to the first flow so the total would be 100 mA – 50 mA = 50 mA.

The 2nd way to think about it is 100 mA must be flowing out of the top of the bottom source.  50 mA of this will be flowing into the bottom of the top source and 50 mA will be flowing toward the resistor junction.   Similarly, at the resistor junction 50 mA will be flowing into the top of the junction and 50 ma will be flowing from the left side with 100 mA flowing out the bottom of the junction.  (Remember Kirchhoff’s current law… it works!)

This was a whirlwind tour through some somewhat complicated ideas.  If anything is confusing, please e-mail me at garyfox@create-and-make.com.

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As always,  thank you for your time.

Gary

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