5/24/16 - AC Power

Today we start with the discussion of average power. We can say we find the average power using an integral divided by the period of that integral. We say this method useless to use, however, as the average value of a sinusoidal function is zero.

We solve for the function given to us on a graph. We know the average value is V_p from the middle of the sine function.

Now we're introduced to the concept of the root mean squared, or RMS value. We define the RMS value to be just the amplitude divided by root 2.

Looking back at the previous class meeting and our definition for power of an AC circuit, we can see that the power is simply the product of the RMS values of voltage and current.

Now we're into AC power, since the voltage and current are usually not in phase, the simple method of multiplying voltage times current won't work for most scenarios.

We do define the product of the amplitudes at apparent power, however, with units of Volt Amps (VA).

We can use this apparent power to find the real power, given in units of Watts, by using a factor called the power factor. The power factor is defined as the cosine of the voltage phase angle minus the current phase angle. We can also find it the power factor if we're given the apparent power and real power.

Professor Mason gives us a basis on why the units are Volt Amps and not simply Watts. He explain it is to prevent the units of apparent power from being confused with units of apparent, or complex, power. He uses the units of torque for example, insisting that the units of torque are meter Newtons, instead of Newton meters like work.

Next, we are given a simple phrase to help us remember which circuit element does what to voltage and current.

ELI the ICE man.

Voltage (EMF) Inductance (L) Current (I) and Current (I) Capacitance (C) Voltage (EMF)

Next, we are given a problem to use our new knowledge of complex power. Given a circuit and it's input signal, we calculate the apparent power and the power factor to find the real power.

As one can tell, our units of real power are simply Watts.. times meters over metes times Joules over Joules times seconds over secoonds times Henry Newtons over Newton Henries times fish over fish times gamma times gamma inverse.

There is also a face in there made out of my traumatized gammas.

Next, we discuss complex power in more detail. We are told that the components of apparent power is the real power plus the reactive power. And we can calculate each component given the phase angles of voltage and current and their amplitudes (or RMS).

Next, we have an example problem. Given the voltage and current functions, we can calculate the phasor of complex power. We then use this phasor to calculate the apparent, real, and reactive power.

We then calculated the power factor and the impedance from the voltage over current.

Javier's peech.. I think. 

Next, we are the voltage RMS and frequncy and a power consumption of 4kW dryer motor. We are told that the current power factor is 0.8 and we want to raise it to 0.95.

We calculate apparent power and reactive powers of the two power factors and use it to solve for the difference in reactive power. We then use a formula to find a value of capacitance to meet our requirement.

Next up for lab, we have two 1k resistors and a 1mH inductor hooked up in series. We will be feeding it different frequencies of signals and observing its current.

First we do 5 kHz and solve for the power dissipated, as well as the voltage and current functions.

We can see that there is about a Pi / 3 phase shift as predicted in the calculations.

And here we have the measured RMS values of the current and voltage, which are about what we caluclated. The top value is supposed to be mA/

The same calculation with a much higher frequency.

Again, our values are close and the top value is supposed to be in mA.

Last one, with a much lower frequency.

Again, our values are close, but not on point as expected.

5/19/26 - Oscillators with AC Op Amps

Today we begin by discussion what happens if we use an AC source on an inverting amplifier. We use nodal analysis to solve for the output voltage as a phasor.

We had a short discussion about frequencies and phase shift, it moved to discussing the wavelengths of a microwave and we were asked to calculate for the wavelength of a 60Hz wave.

 We were solving for something...I forgot exactly what it was since I lost the picture of the original problem.

Next, we have the Relaxing Amplifier. Given an DC power supply, the circuit uses no input single to generate a square wave.

Given the last three non-zero digits of my student ID, we take that to be the frequency to solve our values for. We're given that the period of the relaxing amplifier is some shmoo, and the beta is given as other stuff. We take our R1 and R2 values to be the same so we can simply plug in the natural log of 3 into the equation. Moving everything over the one side yields roughly about 500 ohms of resistance.

Since resistors don't come in 500 ohm values, we just hooked up two 1k resistors in parallel.

We set up our circuit in every circuit to observe it's outcome. Given a bit of noise, that's all the circuit needs to jump start its alternating signal.

Hooked up to the analog discovery, we measured the output voltage as a square wave and the period to be approximately 1.3ms, which translates to 769Hz. Given the uncertainty of our components, we accept this result as a success.

Next, we move into AC Power. Here we begin with the equation for the product of two cosine functions, and since we know the voltage and current to be given as a sinusoidal function, we can relate this to find AC power.

Lastly, we are given this circuit to solve for the currents. Using mesh analysis, we find the currents and then use the current to find the voltage. Since only real part of the impedance dissipates power, we use that to find the total power dissipated in the circuit.

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.