5/17/16 - AC Circuit Analysis

Today we list all the methods of circuit analysis we know for DC circuits, we are told that all the methods can apply to AC circuits as well.

Here we have a problem that we solve for using nodal analysis, we convert everything into the phasor domain and set up our equations as usual, then solve for the voltages at our essential nodes.

Next up is mesh analysis.

Given the circuit already in the phasor domain, we set up our equations to solve for the three currents. We already know one due to the current source in one of the loops and use it to simplify our systems to solve.

Next we have our lab activity. We hook up a resistor and inductor in series and observe the input voltage vs the current and voltage across the inductor as a function of time.

We're told that the cutoff frequency for this circuit is the resistance over the inductance and are told to feed an input of the cutoff frequency, 10 times, and 1/10 of it.

We calculate at the cutoff frequency, the phase shift will be exactly 45 degrees on the inductor voltage.

Here we can see that phase shift.

Next up is 1/10 of the cutoff frequency. We calculate a almost 90 degree phase shift.

Here we can see that there is a approximate 90 degree difference since as the input function comes to zero, the inductor voltage almost reaches its peak.

Next is 10 times the cutoff frequency. We calculate almost no phase shift.

As you can see, the two voltage functions are almost in phase.

Another mesh analysis problem, we solve i_1 and i_2 since i_3 we already know.

Lastly, we can see that we have source transformations as well. They follow the same principles as our DC source transformations except in the phasor domain.

And a small compilation of our drawings.

5/12/16 - Introduction to Impedance

Today we begin the discussion with the concept of impedance. We learn that it is the AC equivalent to DC resistance and is made of a real and jimaginary component.

We are also given the definition of the impedance for resistors, capacitors, and inductors.

Since we know impedance is the sum of the real and jimaginary components, we can covert it into a phasor using the root sum square and the angle from arch tangent.

Next, we are asked to find the current across the circuit as a function of time.

Since we know from Ohm's Law that the current is the voltage divided by the impedance, we can use our phasor operations to find the current and convert it back into a function of time.

As you can see by the chicken on the board, we had a visit from our neighboring lab group.

Here we have another problem, but with a current source. We are asked to find the voltage across the circuit as a function of time and do so again with Ohm's Law. We solve for the total impedance and multiply it by the current phasor and then convert back into the time domain.

We also have another visit from our other neighboring lab group.

A small complication of their controbutions that did not make it into the final p

Now we begin our lab of observing current versus voltage across a circuit with reactive components.

Here we have a resistor hooked up in series with first, another resistor, then an inductor, and finally a capacitor.

The voltage and current plotted against each other for the purely resistive circuit.

As you can tell, there is no phase shift due to there not being a jimaginary component.

The current is in yellow and the input voltage is in blue.

Here we have inductive circuit since the voltage leads the current, or current lags the voltage.

And here we have the current in yellow and the input voltage in blue, we can tell this is the capactive circuit due to the current leading the voltage.

Our measured values for the circuit elements.

The wall of calculations for our currents based on voltages as a function of time.

Next in class, we are given this circuit to analyze. We solve for the impedance of each element and since we know that impedance is the resistance of AC circuits, we can calculate our equivalent impedance using the same rules as resistance.

Here we are given another problem and we use the equivalent impedance to solve for the current, and then use the current to find the voltage across the capacitor as a function of time.

As a finishing exercise, we're told if the resistance equals the reactance, what is the phase angle for the phasor? We can simple do the arch tangent of 1 to find the angle is 45 degrees.

5/10/16 - Introduction to Phasors

Today we start off with a sine function and are given some points. We are asked to find the function of the curve.

Then we give it an offset and find the function with the offset.

We're given our first introduction to the phasor domain here. Since we know that the sine and cosine functions differ by a 90 degree phase shift, we can set up our phasor domain axes as such and then combine the two vectors. We know that their combined magnitude is the root sum square and the angle is arc tangent of the vertical vector over the horizontal vector.

We have the formula for addition of angles in a sine function (which didn't help us for this problem) and voltages given as sine functions. We are told to find the resultant vector and draw it out on the board.

Here we define a phasor, we can say that its components are a real and jimanginary number. The phasor magnitude is simply the root sum square of the real and jimanginary magnitudes and the angle is the arch tangent of the jimaginary over real.

General rules for phasor operations.

Here we attempt our first problem with phasors, we know it's easier to convert them into rectangular and then add, then convert back into polar. As noted by our comment, it only took three tries to attain the correct answer.

There is also a small drawing of a guillotine.

Now applying this knowledge to circuit analysis, we can convert a current or voltage function of time into the phasor domain, but we lose the frequency in doing so.

Another example, we can see that multiplying by the jimaginary results in a 90 degree phase shift and we calculated a phasor from the current given in complex form. We almost had the answer, but didn't account for what region the phasor was in. As you can see by our facial reaction inscribed with detail, we were fairly disgruntled.

Another addition problem, making note of the sin, we modify our functions accordingly and convert to add.

Last problem, we are asked to find the the current across and inductor and are introduce the concept of impedance. We're told that the impedance of an inductor is defined as the jimanginary factors multiplied by the angular frequency and inductance.