The simple circuit analysis techniques we’ll cover in this post are basic, yet important.

Anyone can copy a circuit or project from the Internet or a book. But if you want to be able to design your own circuits, understand how things work, and be able to competently troubleshoot when things go wrong (and they will!) an understanding of the basics of electronics and electricity is crucial.

For example, consider the mega-popular Raspberry Pi. By itself, it can do little besides maybe teach you something about Linux and programming. But if you really want to get the most out of it (or any other micro or platform) you’ll need to connect other things to it.

Not only can learning the basics shed some light on how your platform or board of choice works (and maybe keep you from frying it), it can shed light on how the circuits you connect to it work.

And there are many instances where you may be able to skip the microcontroller all together, using discrete components or ICs to accomplish your task. This can reduce the complexity, cost, and time required to bring your creation to life. Save your $25 Arduino for a more complex project, such as home automation. Of course, one needs to understand at least the basics of electronics to do this.

**Circuit Analysis 101**

This article covers basic circuit analysis, so in that spirit we’ll only be talking about DC circuits. Perhaps we’ll cover AC circuits in another post.

Also, we’ll forget about things like capacitors and inductors for now and focus on circuits containing only resistors in an effort to keep things simple. Again, these types of circuits will probably make an appearance in another post.

Note, however, that most of the techniques I present here will work on circuits containing capacitors and inductors as well as resistors. The only exception would be the way capacitors add up in series or parallel.

If you’ve been dabbling in electronics for more than 5 minutes, you likely know how to combine resistors. However, some of you may be brand-spanking new, so we’ll quickly review this. I will assume, however, that you know the difference between series and parallel circuits and how current and voltage behave in each.

Resistors in series simply add up. For example, the total resistance in the circuit from figure 1 is simply:

R_{total }= R1 + R2 + R3 + R4

*Figur**e 1: series connected resistors.
*

The total resistance in a parallel circuit is always lower than the lowest-value resistor. So, if R1 were the smallest resistor in the circuit in figure 2, the total resistance would be less than R1.

To get the total resistance in a parallel circuit, we add up the reciprocal of all the resistances and then take the reciprocal of this sum.

*Figure 2: resistors in parallel.*

Consider the circuit in figure 2. The total resistance would be:

As we’ll find out in the next post, there are shortcuts that work on *some* parallel circuits.

Many circuits you’ll encounter in real life are series-parallel circuits which are a combination of the two types we just spoke of. An example of a series-parallel circuit is given in figure 3.

*Figure 3: a series-parallel circuit.*

In figure 3, we see that R1 and R2 are in parallel while R3 and R4 are in series. The series combination of R3 and R4 is in parallel with R1 and R2.

To combine resistors in this case, we would work right to left and simply determine which resistor combinations are in series and which are in parallel and use the appropriate formulas to get an equivalent total resistance.

**Ohm’s Law**

Ohm’s Law states that *the voltage across a resistor is directly proportional to the current flowing through that resistor*. Another way to word it is *current flow in a circuit is directly proportional to source voltage and inversely proportional to the resistance of the circuit*. It shows a linear relationship between voltage and current.

This is one of the easiest to remember and most basic rules in electronics. There are two nifty charts associated with Ohm’s Law – remembering them will make your life easier.

*Figure 4: Ohm’s Law.*

The first one is given in figure 4 where V is for voltage, I is for current, and R is resistance. Note that sometimes you may see voltage represented as E rather than V.

This chart is easy to remember and tells us how to solve for any of the three values as long as two are known.

Want to know the voltage? Multiply I x R. Need the current? Divide V by R: V/R.

Here’s a handy trick, take what ever parameter you need to solve for (i.e. voltage, current, resistance) and cover that with your thumb. That will give you the formula you need.

For example, if I want to know the voltage, I’d imagine myself covering the V with my thumb leaving me to multiply I x R. Similarly, if I wanted current, I’d cover the “I” and divide V by R.

Another easy to remember version of Ohm’s law concerns power in a circuit and is given below in figure 5.

*Figure 5: Ohm’s Law with power.*

Notice figure 5 uses E for voltage instead of V. You can use either as your mnemonic, but I find PIE easier to remember than PIV.

We use this triangle the same way we use the other one. For example, if I need to know current and know the power and voltage, I simply divide the power by the voltage: I = P/E.

Using these two figures, one can come up with many different formulas for Ohm’s Law depending on what parameters we know or don’t know about a circuit. The complete version is shown in figure 6.

*Figure 6: all the formulas that one can derive from Ohm’s Law.*

You could memorize figure 6 (or post it at your bench), but any of the formulas above can be easily derived from figures 4 and 5, which are much easier to remember.

Let’s take P = I^{2} x R as an example.

From figure 5 we can easily see that P = I x E (or P = I x V; take your pick).

But, figure 4 tells us that E = I x R.

Substituting that into the first equation we get: P = I x (I x R) which gives P = I^{2} x R. Easy peasy.

According to Ohm’s Law, power, voltage, current, and resistance each have three ways one can solve for them.

The important thing to remember is that regardless of which of the three formulas you choose to use for solving for the parameter you need (i.e. voltage) is that all three formulas should produce the same answer. So, it does not matter which one you pick as long as you have the other two values you need to use for whatever of the three formulas you choose.

So, if I need to find the power and I know the current and resistance, I obviously need to use P = I^{2} x R since I do not know the voltage.

You can use Ohm’s Law to analyze a variety of simple circuits. Let’s do a few examples.

__Ex 1:__ How much current flows through the resistor in figure 7?

*Figure 7*

To get the answer, we remember (using our handy mnemonic from figure 4) that the total current in a circuit like this equals the voltage divided by the resistance:

I = V/R = 9/3,000 = 3 mA.

This seems like a silly example, but real batteries exhibit an internal resistance (though it’s usually less than 3 kΩ!) and this sometimes must be factored into your design.

Let’s do another, more practical example you’re almost sure to see.

__Ex 2:__ LEDs don’t like too much current and should always have a resistor in series with them. How much current flows though the LED in figure 8? Assume a 5 V source, a 0.7 V drop across the LED, and that R = 120 Ω.

*Figure 8*

Since this is a series circuit the same amount of current will flow through the resistor and the LED. Because 0.7 V drops across the LED, we know that 4.3 V drops across the resistor. Using this information, we get the current through the LED:

I = V/R = 4.3/120 = 35.83 mA.

Note that – depending on the LED you pick – you may need to use a larger resistor to keep the current within the LED’s limitations. Ohm’s Law to the rescue again…

__Bonus question (feel free to answer in comments):__ how much power does the resistor dissipate? Hint: eat your PIE…

**Circuit Analysis and Kirchhoff’s Laws**

Mr. Georg Simon Ohm is not the only one who came up with his own circuit analysis rules. Another scientist – Gustav Kirchhoff – also spent some time observing electricity and came up with his own rules.

Ohm’s Law is great but quickly becomes unwieldy when analyzing all but the simplest circuits. That’s where Kirchhoff’s Laws come in. The two together form a powerful suite of tools for analyzing more complex circuits.

Mr. Kirchhoff is famous for two laws: Kirchhoff’s Current Law (KCL) and Kirchhoff’s Voltage Law (KVL).

**Kirchhoff’s Current Law or KCL **

Kirchhoff’s Current Law states that *the algebraic sum of currents entering a node is 0*.

To put it another, perhaps easier to understand way, it says that *the sum of all currents entering a node is equal to the sum of all currents leaving a node*. After all, electrons don’t simply evaporate and disappear!

Figure 9 illustrates this.

*Figure 9: Kirchhoff’s Current Law.*

Here we see I1 and I2 entering node 3, while I3 is leaving the node.

It’s easy to see that I1 + I2 = I3 or we can write it as I1 +I2 – I3 = 0, which shows us that the algebraic sum does indeed equal 0.

Let’s do an example problem using KCL.

__Ex 3:__ Find the battery current I_{B} in figure 10.

*Figure 10*

Using Kirchhoff’s Current Law at node 1 gives us:

I_{B} = I_{1} + I_{2}

Ohm’s Law tells us that:

I_{1} = V/R = 20/5 = 4 A

And, since the resistors are in parallel, 20 V shows at both:

I_{2} = 20/15 = 1.33 A

Therefore:

I_{B} = 4 A + 1.33 A = 5.33 A

**Kirchhoff’s Voltage Law or KVL**

Kirchhoff’s Voltage Law states that *the algebraic sum of all voltages around a closed path or loop is 0*.

To put it another way, *the sum of all voltage rises around a closed loop equals the sum of the voltage drops around the loop*.

We can also phrase it as *the sum of the voltage drops is a series circuit is equal to the total voltage applied*.

Figure 11 illustrates this.

*Figure 11: Kirchhoff’s Voltage law.*

Because figure 11 is a series circuit it is a closed loop and the same current I flows through all the elements.

Using KVL, we can write: V_{s} – VR_{1} – VR_{2} = 0 or we can rearrange and write: V_{s} = VR_{1} + VR_{2}

Both say the same thing.

Let’s do a quick practice problem.

__Ex 4:__ How much current flows through the circuit in figure 12?

*Figure 12*

Those who have taken engineering courses may recognize this type of circuit as being a fictitious, engineering-esqe circuit which exists only on paper to mess with your mind. You can always tell these because they usually have at least two off-the-wall sources and are either impossible to build as shown or never encountered in real life.

Anyway, the direction of current flow is already given as clockwise – this matters. The polarity of the voltages also matter. The polarity of the resistors in this fictitious circuit are given to you. Normally, we assume voltage to drop across a resistor. For figure 12 we’ll treat voltage rises as positive and voltage drops as negative as we follow the path of the current.

If the current’s path is not given in a situation like this (which is likely), I like to assume it travels clockwise, but you can pick counter-clockwise, too. You *must* be *consistent* with your method of choice or you won’t get the right answer. For more about this, see Conventional Current vs. Electron Flow: Which is Correct?

Using KVL, we can write:

20 – V_{1} – 50 – V_{2} = 0

Ohm’s Law tells us that:

V_{1} = I x R = I x 1 = I

And

V_{2} = I x 2

So, 20 – I – 50 – 2I = 0

Rearranging algebraically, we get:

-I – 2I = 50 – 20 which after simplifying is -3I = 30

Solving for I, we get -10 A.

The negative sign tells us that the current is actually flowing the opposite direction we assumed it was, since it is impossible to have a negative number of electrons. If we were to draw the arrow representing current the opposite way and redo our analysis correctly, our answer would be 10 A. Notice that the absolute value (or the amount) of the current that flows is the same either way.

__Bonus question:__ given resistors with the right power rating and voltage sources are available, is it possible to build this circuit with the exact parameters shown in the figure? Why or why not?

**More Circuit Analysis Tricks to Come**

As usual, this post has grown into a 2,000+ word monster, and there’s a lot of circuit analysis tricks we haven’t covered.

In part 2 of this series we’ll look at some other tricks and applications to extend the things we learned here.

We’ll look at handy tricks like voltage division, current division, and superposition. Then (perhaps in part 3) we’ll get into some more advanced methods like nodal and mesh analysis and also Thevenin and Norton.

Truthfully, one can easily devote a whole post to each of these advanced techniques. We’ll see about that when the time to discuss each comes.

Meanwhile, comment and tell us about your favorite circuit analysis trick or technique.

Phil says

Looks to me like the polarity on the resistors in figure 12 would actually be reversed.

Brian says

What does everyone else think? C’mon everyone, take a stab at this.

Lucian says

A theoretical opinion with the same results.Figure1 containing one source with the electromotor force(difference of potential) E[V;volts],the resistors R1,R2,R3,R4[ohms] passed by the same current with the intensity I[A;ampers].The drops of potential on each resistor will be,according to the law of Ohm,V1[Volts]=I[Ampers]*R1[ohms],V2=I*R2,V3=I*R3,V4=I*R4.According to the 2nd law of Kirchhoff the electromotor force E is equal with the summ of the drops of differences of potential V1,V2,V3,V4: E=V1+V2+V3+V4=I*R1+I*R2+I*R3+I*R4=I*(R1+R2+R3+R4).But,according to the Ohm’s law E=I*Re,Re=equivalent resistance of the resistors R1,…R4 in serial connection;that implies I*(R1+R2+R3+R4)=I*Re which leads to R1+R2+R3+R4=Re.