Combining Voltage Sources

Icon to return to the previous section.
Icon to return to table of contents.
Icon to go to the next section.

Part 1: Fundamental Concepts: Combining Voltage Sources

So far, we have looked at what happens when we combine resistors in different circuit configurations.  In many cases, there was a voltage source involved and the currents and voltages in the circuit were investigated.  What about voltage sources, can they be combined?  The answer is, yes.  Let’s start by looking at what happens when voltage sources are combined in series.  This is shown in the following figure along with a resistor acting as a load.

Voltage sources sonnected in series.

Figure 1. Voltage sources connected in series.

The concept of combining voltage sources should be a familiar one.  In the pictorial representation of a flashlight,  there are two batteries shown.  If each battery is 1.5V, we intuitively understand that the two batteries together provide 3V.  Why does this work?

A voltage source, along with real world examples such as a battery, provide the potential to do work.  From an electrical theory standpoint, they are not the source of the charges in a circuit, they are what provides the potential that causes the current to flow.  Ideally, while doing this, they also have another useful characteristic.  They internally have no resistance to current flow.  In actuality, those things that act like voltage sources, a battery for instance, do have some internal resistance but it is low enough to be neglected in many cases.

Take a look at Figure 1 with these two characteristics, that a voltage sources maintains a potential difference and has no resistance, in mind.  Begin by looking at points “a” and “b”.  The potential at point “b” is 5V higher than the potential at point “a”.  This is because a voltage source maintains a potential difference across its terminals.  The same reasoning applies to points “b” and “c”.  The voltage at point “c” is 10V greater than the potential at point “b”.  What is the potential between points “a” and “c”?  It is 15V.  This occurs because the 10V voltage source is maintaining its voltage difference in reference to the already elevated voltage at point “b”.

The current flowing in the circuit is governed by Ohm's law.  In this case, I = 15 ÷ R1.  How does the current that is returning to point “a” in the circuit make its way back to the 10V source?  It does this by flowing through the 5V source.  Remember that voltage sources have no internal resistance — they look like a piece of wire.  You can think of a voltage source as a piece of wire that magically maintains a voltage difference between its ends.

What about connecting voltage sources in parallel?  This bad idea is shown in Figure 2.

Figure 2. Don’t do this!

Voltage sources are an abstraction that are used during the analysis of circuits.  With the properties we assign to them of always providing a potential with no internal resistance, the schematic shown in Figure 2 will result in a huge (theoretically an infinite) current.  There is a 5V difference between the two voltage sources with 0Ω resistance in the circuit. Ohm's law gives us 5V = I * 0 or I = 5 ÷ 0!  You should never connect voltage sources of different values, or their equivalent in real world devices, in parallel.

That said, it is fairly common practice to connect batteries in parallel, positive to positive and negative to negative, to increase the overall capacity of the battery pack while maintaining the same voltage.  This works because batteries, while convenient to think about as voltage sources, are far removed from them.  They are not ideal voltage sources.  When connecting batteries in parallel they should be of the same type (same part number, age, etc.) and have the same voltage across their terminals before the connection is made.  If they are not, you will end up with a version of the problem shown in Figure 2 because of the mismatch.

As a side note, it is common practice to place two automotive batteries in parallel (positive to positive and negative to negative) when jump starting a car with a dead battery.  In this case, the mismatch between the two batteries results in a current flowing from the charged battery to the dead battery.  This is the desired result.  The resistance in the connections, batteries, and jumper cable wire limit the current flow.  It is, of course, a good idea to have one car running while you do this so that you don’t end up with two dead batteries.  Current will flow from the charged battery into the dead battery until they are both at the same voltage, allowing you to start the car with the dead battery.

With the knowledge that parallel connections of two voltage sources is a bad idea, let’s return to the investigation of series connections.  The figure below illustrates a slightly more complex set of connections involving three voltage sources.

Circuit with three voltage sources.

Figure 3.Circuit with three voltage sources.

The circuit shown in Figure 3 has three branches.  On first inspection, you might think that the 6V source is in parallel with the other two series connected voltage sources, but it is not.  There is a resistor, R2, that is series with the 6V source limiting the current.  The goal is to find the branch currents I1 and I2.  The total current, which is not labeled, is the sum of these two.  To aid in the analysis, the figure below is the same as Figure 3 with some additional annotations.

Annotated circuit with three voltage sources.

Figure 3.Annotated circuit with three voltage sources.

All of the voltages indicated in Figure 4 are in reference to the common connection, or node, between the negative terminals of the 5V and 6V sources. To determine the current though R2 (I1) and R1 (I2), we need to know the voltage present across them and their resistance.

The voltage at the top terminal of R2 is 15V because the 10V and 5V sources are connected in series, resulting in a 15V difference between our common measurement point and the positive terminal of the the 10V source.  The bottom terminal of R2 is 6V in reference to the same common connection point.  The voltage difference across R2 is therefore 9V (15-6=9).  Using Ohm's law yields:

            V = I * R            ;Ohm's law

            I = V/R              ;Rearrange

            I = 9V/1000Ω    ;Put in R2 and voltage

            I = 0.009A         ;The current through R2

Finding the current flow through R1 can be done directly using Ohm's law as there is simply 15V across its terminals.  Substituting the circuit values into Ohm's law yields: 15V = I2 * 2,000Ω or I2 = 0.0075A.

Key Concepts

• Voltage sources act in circuit like a piece of wire that has a potential difference (voltage) across its ends.

• Connecting voltage sources, or things that act like them, in parallel is a bad idea.

• Batteries can be connected in parallel with the caveat that they be of the same type and the same voltage.  The batteries internal resistance helps keep any imbalance in the batteries voltage from resulting in too large of a current flow.

• You can have multiple voltage sources in a circuit as long as they are not directly connected in parallel.

Icon to return to the previous section.
Icon to return to table of contents.
Icon to go to the next section.

copyright © 2021 John Miskimins