Software & Electrical Engineering Tutorials

Mesh Current

July 23, 2009

3:43 am

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: Mesh Current Loops

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…

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

Solving the 3×3 system

I1 = 5 A

I2 = 2.5 A

I3 = 2 A

Voltage Dividers

by Jeff Schuler

July 21, 2009

12:03 am

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. Series Circuit 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.

$V1 = Vs( \frac{R1}{R1 + R2})$ $V2 = Vs(\frac{R2}{R1 + R2})$

Series Resistance Circuit

The general equation for a voltage divider is given below, where V_o is the measured voltage, V_s is the source voltage, R_o is the resistance across which the voltage is measured, and R_T 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 R_o 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 R_o 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, V_o is measured across resistors R₂ and R₃ only. R_o only includes the resistance between the plus and minus of V_o.

$V0 = Vs(\frac{R0}{Rt}) = Vs(\frac{R2 + R3}{R1 + R2 + R3 + R4}) = 20(\frac{2k + 3k }{1k + 2k + 3k + 4k}) = 10V$

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.

Adding Parallel Resistors

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 R_o 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

by Jeff Schuler

July 20, 2009

2:19 pm

Engineersphere.com

What Is Current?