Scroll 5

Resistor Networks

(This information applies several of the problems in this scroll) This worksheet and all related files are licensed under the Creative Commons Attribution License,version 1.0. To view a copy of this license, visit http://creativecommons.org/licenses/by/1.0/, or send aletter to Creative Commons, 559 Nathan Abbott Way, Stanford, California 94305, USA. The terms andconditions of this license allow for free copying, distribution, and/or modifcation of all licensed works by

the general public. Resources and methods for learning about these subjects (list a few here, in preparation for your research):

Summary:

To avoid making the TOPHEAVY BLUNDER. Remember

the following Kung Fu Calligraphy. For current division, which occurs in parallel circuits, Req is on top. For voltage division, which occurs in series circuits, Req is on the bottom.

Minion Master Kirchhoff “There are some mysterious circuit components whose values depends on a value somewhere else in the circuit. These components are called dependent sources. Look at the image to the left and identify the two dependent sources. One is a current dependent source (CDS) and the other is a voltage dependent source (VDS).

See how the 5Vd VDS depends on the voltage flowing through the 3Ω resistor, but the voltage produced by the 5Vd VDS is 5*Vd. The same happens with the Ix CDS. Find the following values.”

Vd = ___________ Ix = ___________ V1 = __________

Master Kirchhoff “Remember that you can reduce a resistor network by combining resistors in series and in parallel. Remember that resistors in parallel are connected at both end while resistors in series are only connected at one end. Looking at the circuit on your left see how the 2Ω, 3Ω, and 4Ω resistors are in series, and these combined are in parallel with the 9Ω resistor.

“Find the resistance equivalent seen from the terminal A and B. Req = __________

HenchmenMaster Kirchhoff “Current division will be a powerful skill to learn as you master circuits. Look at the circuit on the left and convince yourself that as current moves through a resistor the current does not change, but there is a change in voltage. Thus the current flowing through the point x is also 12A. But we cannot say that the current flowing through the 6Ω resistor is also 12A since the current is divided into two branches shortly after point x.

I will now guide you through a few steps to help give you further insight into how current division works. First find the equivalent resistances of the two resistors in parallel. Req = ____________

Your circuit should look like the one on the left. Notice that your I-equivalent (Ieq) is also 12A since it is in series with the 12A source. Using ohms law, find the voltage across Req = _____________. According to how I referenced ground, the voltage at point x is the same as the voltage across Req since it is the only element between x and ground and ground is at 0 volts. This shows that the 6Ω and 3Ω resistors are two paths down the mountain across the same distance. Or the voltage

drop across the Req resistor = ΔV6Ω = ΔV3Ω. Using Ohm’s law find the current through the 6Ω resistor.I = _____________

Lets look at the math. V Req=I eq∗Req; Req=R1∗R2R1+R2

(for resistors in //); ΔV Req=ΔV R1=ΔV R2;

Thus IR1=ΔV Req

R1= R2R1+R2

∗I eq.

Using this equation verify your answer for I. I = ____________

Answer = 4

Master Kirchhoff “Now let’s explore voltage division one step at a time. Voltage division relates the current between resistors in order to find the voltage drop across individual resistors. The first step is to simplify the resistors as you can see below. Find Req2 of all of the resistors as seen from the voltage source. Req2 = _______________

Figure 1Figure 2

See how Ieq is the the same current flowing through the 1.6Ω resistor, Req1, and Req2.Looking at figure 2 determine Ieq by using Ohm’s law. Ieq = ___________Looking at figure 1 determine V1 by using Ohm’s law. V1 = ___________Remember that the voltage across two parallel (//) resistors are the same.

Let’s look at the math. I eq=V s

R1+R2; ΔV R 1=Ieq∗R1; thus ΔV R 1=

V s

R1+R2∗R1

With Req1 = R1 and 1.6Ω = R2Using the equation verify your answer for V1. V1 = _________________

Answer = 7.2

Master Kirchhoff “Remember that current division is to find the currents in two // resistors, and voltage division is to find the voltage drops across two resistors in series.”

Boss

Master Kirchhoff “Using everything that you have learned find Vy.”

Vy = ___________

(Hint) Master Kirchhoff “Two things in // have the same voltage across them. Notice how the 7Ω and 11Ω resistors are in // with the 54V voltage. This indicates ΔV7Ω + ΔV11Ω = 54 V. Use voltage division to find Vx.”

(Hint) Master Kirchhoff “Notice that the same current that flows through the 13Ω resistor is the same current flowing through the 7Ω resistor. Find the Req of the two resistors in series and then use current division to find Ia.”

(Hint) Master Kirchhoff “Vy = 9*Ia – 54V.”

Practice Problems:

Minions

Henchmen

Bosses