Note: This is modified archive of the original class webpage which can be found using the Wayback Machine.
In discussion we went over passive sign convention and determining polarity for components on a circuit. Given a circuit, first mark the direction of the currents. When a current source exists, this is simple, but for other cases you can make a guess as long as you are consistent in each branch. From this determine the polarity of every component. For resistors, current enters the positive side. For voltage sources, polarity is given and for current sources you may guess the polarity. It does not matter whether your guess is right or not, because the sign of your answer accounts for this.
Once the circuit is fully annotated, apply KVL and KCL as needed. Use the question "what do I know about this circuit?" to guide you, and solve the resulting system of equations. In the example shown, we applied an equivilant resistor in the center then used KCL at the top center node and KVL in the right most loop. For KVL account for all voltages by the first sign that appears (add for '+' and subtract when you see a '-') going clockwise.
For determining power, apply the passive sign convention by following the direction of the current into a component and multiplying the current by the voltage (P=IV) or an equivilant form. Note that current entering a positive terminal is positive power and current entering a negative terminal is negative power by the passive sign convention.
Delta to Wye Transform Example
We went over an example of the node voltage method similar to problem four from the homework. To apply the node voltage method, first find the reference node which best appears to link all other nodes together. From here label all other essential nodes and define the corresponding voltages between the node and reference. Then define all the currents associated with each essentail node and note resistor polarities accoringly. Using KVL find an expression for each current and apply KCL at each node except the reference. You should derive enougth equations to solve for the voltage at each node.
During office hours, I realized that my procedure for KVL is different from Professor Ayer's. I perform KCL in a clockwise loop and using the first polarity symbol reached at each component. Similar to how passive sign convention records disspiated power as positive I account for voltage drops (with resistive components being positive in the equation). I also go through the full loop using a variable for the current, and then solve for it. Ayers uses a shortcut to write out the currents at once; however, I believe my method may be less prone to mistakes.
Super Node Notes (1) -- and known node voltage
Quiz 1 Solution : NVM (Recomended)
OpAmp 1 (Problem)
To solve an op amp circuit there are two general principles to follow: First, no current enters either input terminal of the amplifier. This is called infinitie input impedence (III). Second, during linear operation, Vn=Vp. This is the virtual short (VS), which is only valid where |Vout| < Vcc. Otherwise the output voltage is saturated and fixed to +/- Vcc.
My recomendated procedure is as follows:
We went over an AC steady state analysis example and working with complex numbers. Generally, you want to convert each component to it's corresponding impedance, and then solve in the phase domain using the same techniques you learned before exam 1.
The challenge here is complex numbers. You will need to be efficent at converting between polar and rectangular form, and recongize that addition/substraction are often easier in rectangular form, and multiplication/division are often easier in polar form.
Like before, you can use a calculator to solve the systems; however, a TI83/84 will not solve matracies with complex numbers. In many cases, you can use node voltage method to derive fewer equations and solve by hand. Take advantage of the calculator's ability to perform arithmetic on complex numbers (make sure to use parenthesis as needed). If necessary, you can use the trick Professor Ayer's sent.
For homework problem number two, I tried going over this and missed a "j" and subsequently added it back in the wrong place. With that being said, the way to solve this is to simplify the equivilant impedance enough to group all the imaginary terms and set it to zero. Be very careful when doing complex arithmetic, because it is easy to make mistakes.
The maximum power problem is much like we've seen before in DC analysis. First do a thevenin analysis on the circuit, then the load impedance will ideally be the conjugate of Rth, as shown in the notes.
Regarding complex power, it may seem confusing with all the types of power. Much like impedance, complex power (S) is made up of the real thing - what you worked with before (think resistance) in addition to something else (think reactance). The average power (P) makes up the real component and reactive power (Q) is the imaginary component (S = P + jQ ). Often, it is desirable for a balanced power source where the total load is resistive only (power factor is one). This is because, reactive power draws current even though it does not disspiate energy over a fully cycle. This can result in strain on the grid.
Reactive power is best understood as a temporary drawing of energy by a load which will be returned later. Capacitors temporarily store energy in an electric field and inductors do so within a magnetic field, but release such energy in the other half cycle. As this article by Peter Sauer describes, it is much like carrying a bucket of water up a ladder and back down: no work was done. Instead reactive power only affects the phase difference between current and voltage.
