## What Is Current?

Let’s start with charge.  Charge is measured in coulombs, 1 coulomb has $6.24 x 10^{18}$ electrons.

Charge of 1 electron:  Qe = $-1.6 x 10^{-19}$ coulomb.

Charge is carried by negative electrons and/or positive protons.  In a solid, the current is carried by electrons.

Current: is the rate of charge flow.

Current (I) = $\frac{Charge}{Time}$ OR I = $\frac{Q}{T}$

1 Ampere = $\frac{1 coulomb}{1 second}$

1 A = $\frac{1C}{1 S}$

Amp = $\frac{C}{ S}$

## Mesh Current

Here I will show you how to calculate the different currents in each loop of the figure below using the mesh current method.

This is my favorite approach to a problem like this one: 1) Identify meshes in a planar circuit.

2) Identify currents unknown.

3) Write KVL for each mesh.

4) Simplify and Solve.

Easier said than done I suppose, so let’s do it:

It is important to understand how many equations you are going to need.  Here is how you do that:

Using Node-Voltage:   #Equations = Ne – 1

Using Mesh-Current: #Equations = Be – (Ne – 1)

Where Ne = Number of Node,  Be = Number of circuit components.

Here we have:

Ne = 4

Be = 6

Be – (Ne – 1) = 3 equations.

Now you are ready to start writing your mesh equations.  You should only do one loop at a time.  This is simply KVL at work here, adding up the voltages around the loop and making sure they sum up to zero or to the source voltage.

Mesh 1: 5 ( i1 – i3) + 26 (i1 – i2) = 80

Mesh 2: 26 ( i2 – i1) + 90 (i2 – i3) + 8 (i2) = 0

Mesh 3: 30 (i3) + 90 (i3 – i2) + 5 (i3 – i1) = 0

We now have 3 equations and 3 unknowns, this is what we want and we should go ahead and move the “80” in the Mesh 1 equation to the left side so that all equations sum to zero.

Confused by this? “( i2 – i1)”

That is normal, and probably the most confusing concept.  All 3 of the currents in all 3 loops are heading in the same direction, but in relation to each circuit component (such as 90Ω resistor), they are opposite.  If you are in loop number 1, this current (i1) is your reference current for this given loop, and the current through a resistor in that loop is equal to that reference current minus the opposing current on that given component.  If the currents were heading in the same direction, they would simply add instead of subtract.  This way we have summed up the total current through each component and can therefore calculate the voltage.

You can either use a TI-89 to solve for these unknowns…

OR $\begin{pmatrix}31&-26&-5\\-26&124&-90\\-5&-90&125\end{pmatrix}$ $\begin{pmatrix}i1\\i2\\i3\end{pmatrix}$ = $\begin{pmatrix}80\\0\\0\end{pmatrix}$

Solving the 3×3 system

I1 = 5 A

I2 = 2.5 A

I3 = 2 A

## Using Voltage Dividers

The voltage divider equation is arguably the most important equation for an electrical engineer to know.  At the very least, it is one of the most fundamental.  Although the voltage divider technique becomes cumbersome when applied to larger circuits, no other method is faster when it comes to finding voltages in smaller circuits.  This beefy voltage divider explanation was kindly donated by Ryan Eatinger. The simplest form of a voltage divider circuit is shown in Figure 1.  V1 and V2 can be found using the following equations. $V1 = Vs ( \frac{R1}{R1 + R2}) = 9 ( \frac{1k}{1k + 2k}) = 3V$ $V2 = Vs ( \frac{R2}{R1 + R2}) = 9 ( \frac{2k}{1k + 2k}) = 6V$
The voltage divider equation applies to series circuits where the current remains constant throughout the circuit.  If current is constant for all resistors, then it can be taken out of the equation.  This is the true advantage of the voltage divider.  If there is a choice between working with currents and voltages and working only with voltages, the choice becomes obvious. Here’s how it works.  Once again, we have the circuit in Figure 1.  The current remains constant throughout the circuit, meaning that the current through the source equals the current through R1 equals the current through R2. $Is = I1 = I2$
Recalling Ohm’s law, write the equation above in terms of voltage and resistance.

Ohm’s Law: $I = \frac{V}{R}$ –> $\frac{Vs}{R1 + R2} = \frac{V1}{R1} = \frac{V2}{R2}$

These equations can now be used to find V1 and V2. $V1 = Vs( \frac{R1}{R1 + R2})$ $V2 = Vs(\frac{R2}{R1 + R2})$ The general equation for a voltage divider is given below, where Vo is the measured voltage, Vs is the source voltage, Ro is the resistance across which the voltage is measured, and RT is the equivalent resistance of the circuit.  Figure 2 shows the corresponding circuit.

General Voltage Divider Equation $V0 = Vs(\frac{R0}{Rt})$

A voltage divider is not always in the simple form shown so far.  Recognizing a voltage divider is a skill that takes time to develop.  This article introduces some variations on the basic voltage divider circuit that you may encounter.  The best way to solve a voltage divider is to simplify it to the basic form shown in Figure 2.  Once in this form, apply the general voltage divider equation to find the desired voltage.

### Series Circuits Applying the voltage divider equation to a series circuit is a fairly straightforward process.  It’s simply a matter of identifying which resistors make up Ro and then adding all the resistors together to find the equivalent resistance.

Example 1: In Figure 3, there are four resistors and you’re trying to find the voltage across one of them.  The resistors are all in series, making the equivalent resistance of the circuit 10 kΩ. $V0 = Vs(\frac{R0}{Rt}) = Vs(\frac{R4}{R1 + R2 + R3 + R4}) = 20(\frac{4k}{1k + 2k + 3k + 4k}) = 8V$

Example 2: When analyzing a circuit, pay attention to the orientation and location of the plusses and minuses.  Voltages aren’t always across one resistor.  In Figure 4, the terminal voltage is across the combination of resistors R2, R3, and R4.  Adjust the Ro accordingly to include the equivalent resistance of the three resistors. $V0 = Vs(\frac{R0}{Rt}) = Vs(\frac{R2 + R3 + R4}{R1 + R2 + R3 + R4}) = 20(\frac{2k + 3k + 4k}{1k + 2k + 3k + 4k}) = 18V$

Example 3: You should also remember that voltages aren’t always measured to ground.  In Figure 4, Vo is measured across resistors R2 and R3 only.  Ro only includes the resistance between the plus and minus of Vo. $V0 = Vs(\frac{R0}{Rt}) = Vs(\frac{R2 + R3}{R1 + R2 + R3 + R4}) = 20(\frac{2k + 3k }{1k + 2k + 3k + 4k}) = 10V$

### Circuits with Parallel Resistors:

Voltage dividers apply to resistors in series.  If you encounter a circuit with resistors in parallel, you must combine any parallel resistors before applying the voltage divider equation. Only after combining the parallel resistors will the voltage divider equation work.  For this example, the parallel combination of R2, R3, and R4 combine to form the Ro in the general voltage divider equation. $V1 = Vs(\frac{R0}{Rt}) = Vs(\frac{R1}{R1 + R1||R3||R4}) = 10(\frac{1k}{1k + 1k}) = 5V$ $V1 = Vs(\frac{R0}{Rt}) = Vs(\frac{R2||R3||R4}{R1 + R1||R3||R4}) = 10(\frac{1k}{1k + 1k}) = 5V$

Well, that should just about cover voltage dividers.  As always, if you have any questions feel free to make a comment or send me or another one of the admins a message and it will be taken care of!  Thanks again to Ryan Eatinger (reatinge@ksu.edu) for the article.

## Welcome

This is the first post at EngineerSphere.com, I’ll consider this a landmark.  The goal of EngineerSphere.com is to bring the technical community together to make things easier on everybody.  There will be frequent informative posts on a multitude of electrical / computer engineer-related subjects.  As time goes on, other focus’ will be added on to better accommodate the entire spectrum.  Whether you are an undergrad, graduate, or even a professor, everyone can find something useful here.  If there is a subject that you would like covered in greater detail send an email to Jeff@engineersphere.com and an admin, a knowledgeable poster, or myself will gladly elaborate on given subject.  Don’t have time for that?  Sometimes we don’t either, but someone reading does, simply post your specific question in the Q & A section of the website and enjoy free help.  If you are bored or need a 5 minute break from that Fourier Transform, log in and answer a few questions!

The ‘Vault’ Section of the website is for specific members only and is available by invite only.  If you would like access to the The Vault, email jeff@engineersphere.com and just let me know.  The Vault will contain a large database of solutions manuals that users have collected over the years.  A separate, public vault will be available to the community that will consist of freeware type software such as p-spice, visual studio, and PCB artist.  Again, if you have something you would like to contribute to the public or private vault, or are looking for something in particular, make it known and it shall be done.

That is all.  Enjoy.