6/2/15 - AC Circuits Cont.

Today, we begin by looking at what happens when more elements are in an AC circuit. Particularly the RLC version.

We state that Z is total rectance of the circuit and we find it by the sum of the square of the resistance and the square of the difference of the inductive reactance and capacitive reactance.

Since we don't have an inductor in the circuit, we just have the reactance of the capacitor.

We calculate the theoretical limits of what would happen if certain circuit elements go to infinity.

We say since that is the equation to find current, we say that current should increase or decrease by a factor of 0 < X < 2 since it is bound by the limit.

Here we hook up a resistor and capacitor in series and apply LoggerPro sensors to the circuit.

The voltage probe was in the wrong position, but was later fixed.

Here we have the potential and current versus time graphs. As shown, the phase shift is present, but is not exactly a 90 degree phase shift.

With the voltage probes still connected incorrectly, we change to the right resistance that we used for the calculations and take new data.

Here we take more data with different frequencies. We notice that the shift is different when we changed the frequencies.

We compare our theoretical values to our experimental values for reactance. We explain the large error due to the unknown resistnace in the function generator. Even with the voltage probes in the correct position, there was still large error.

We then find the displacement of the time between two peaks on the voltage and on the current graphs, find the difference and divide by the period.

We take the arctangent of that and find the displacement angle between the two peaks.

As expected, it was not a perfect 90 degree angle and the higher frequency produced a lower angle displacement.

When we make our theoretical calculations, we compare them to the actual experiential ones and find we're not too far off.

Here we calculate the power dissipated by the circuit, but since only the resistor element in the circuit dissipates power, we can't use the standard formula of current times voltage.

We use the derived formula of current squared times resistance and find power through that.

Here we measure the voltage and current with a multimeter.

Here we took the correct graph with LoggerPro and will now use the data to compare to our collected data from the multimeter.

We learn about the resonate frequcney in the LRC circuits where the capacitive reactance and the inductive reactance cancel out and the circuit is a purely resistance circuit.

We derive that frequency to be 1 / (2\pi * inductance * capacitence).

We calculate a frequency and compare it to our measured frequency from the multimeter.

We hooked up the RLC circuit to the function generator as normal, but then adjusted the dial on the function generator until the current reached its peak. 

We compare this experimental value to our theoretical value and find a percent error. Again, we attribute this error to the unknown resistance in the function generator.

5/28/15 - Into to AC Circuits

We start off the class by looking at how an AC current works. Since it's a sine function, it's value is always changing and finding an average is not as easy as DC.

Since we take the average the sine wave, we can find the average with an integral.

But an integral is a lot of work, so the professor cheats a bit and says we can just use the sine squared plus cosine squared equaling to one, we say that they're equal and equate the square to one half.

Then we solve for the the root mean square as one over root two.

If the wall socket gives a 120 V_rms, we can find the actual peak voltage by multiplying by root2.

Here we hook up a function generator to a resistor and hook up the current and voltage sensors to the circuit. We then record the data in LoggerPro.

Here we can see the voltage and current graphs.

The same setup and graphs, except with a different resistor and frequency. 

Comparing our calculated V_rms and I_rms with our experimental values, we calculate a the small error from the graph and the multimeter.

The large percentage is likely from the low voltage and uncertainty of LoggerPro.

Comparing the voltage and current graphs, we have a linear line sloped at a 45 degree angle. Which would make sense due to the direct relationship between the voltage and current.

The next setup involved a capacitor rather than a resistor.

Since the current is the scalar derivative of voltage, and the voltage is a sinusoidal wave, the result is a scaled co-sinusoidal wave.

Here we see the potential is still a sine wave, but the graph between the potential vs current is now a circular function due to the constant phase shift between the two graphs.

We explain the phenomenon as reactance, and the \Chi for capacitors is equal to one over capacitor times and period. We solved for V_rms given the rectance.

With this setup, we attempt to record the values of the graphs of the AC - C circuit.

Here, we calculate the theoretical values of the V_rma and I_rms and compare them to the graph's values. 

We find that there is a large error value from the graphs, we figure this due from the unknown resistance given by the function generator.

However we see that the phase shift, time taken between two peaks, is roughly one fourth of a period.

This works in accordance with the phase shift from a capacitor.

We repeat our experiment again except with an induction coil in the circuit this time. We derive the voltage for an inductor and define the inductance of a inductor to be the inductance times period.

We then can plug in to find V_rms and I_rms.

We compared the measured to theoretical values and found only a large error in our current. Again, we explain the due to the unknown resistance in the function generator.

5/26/15 - Inductance Cont.

Today, we start off by drawing the curve of current in a R-L circuit. We draw it as an exponential growth curve.

After all the demonstrations, we're given a resistor and induct or to use in our data collection. 

Here is the measurement of the resistor from reading the colour stripes on the resistor.

Given the formula we derived last week for inductance, we calculate our theoretical value for inductance. We also calculate the resistance of the inductance coil since it is made out of copper wire.

Since we have the period from the time constant, we can calculate the frequency of the wave. 

Here we hook up a function generator to the oscilloscope. As shown, we can see the square wave on the oscilloscope.

Here we hook up the resistor in series with the inductor...which was wrong. It produced an interest graph though.

Here we did the calculations for the induction coil we used in the actual circuit.

We found a new resistance and a new time constant.

This time constant is also calculated with the resistance of the resistor in the circuit and the approximate resistance of the function generator.

Given the new time constant, we found a new frequency.

With the correct circuit, we can see the graph on the oscilloscope is showing the exponential inclines and declines.

Zooming in, we can see the curve more clearly, however it is still hard to measure since it is still an analog graph.

When we zoom in and scale the graph, we attempt to make a measurement for the period of the wave.

Using the half time constant formula, we calculate an approximate time constant.

We multiply the time constant by the resistance and find an approximate inductance for our circuit. We then solve for the amount of loops in the inductor and find a value with a large error.

We estimate this error is due to the analog graph and uncertainty of the resistance in the function generator.

Given an Active Physics web app again, we answer some questions on the demonstration. We realize that we've already done the same one before and move on. Thankfully we move on, since our app crashed.

The last thing we observe in class is a R-L circuit.

We observe the top and middle branches' inductance time constants.

Since we can find the max currents, we can find the current through the circuit at any given time.

The middle branch only has an inductor, so that branch is instantly at max current.

The top branch we find the current based on the max current at a time after the switch is closed.

We notice that the voltage across the middle resistor is, as expected, 45V; but the top branch has a potential difference drop across the resistor that is lower than the max due to the inductor.

Now given a potential difference, we're asked to find the time it takes to reach that potential difference.

 Since the voltage of the resistor is the max voltage minus the inductor's voltage, we find that potential difference across the resistor. Since we have the current current flowing through the branch, we can back substitute into the formula and solve for the time.

Finally, we solve for the energy stored in the inductor with the formula of half inductance times current squared.  We notice it has a very small amount of energy stored.

5/19/15 - Intro to Inductance

We begin the class with an Active Physics Java applet. It allows us to have an dynamic interaction with a magnetic field and a plate. We can control how fast the plate rotates and we have a flux and or emf versus time graph below.

Here we just answered the question from the applet's sidebar. Given certain actions for the plate's rotational frequency and direction, we can see that the flux is proportional to the area and the emf is related the to rotational frequency.

Here we see some key points on the graph, and conclude that the emf is related to time derivative of the flux. And the magnitude of the emf increases and frequency increases.

Here we have the large horseshoe magnet and a copper pipe on two conductive rails, a current flows through the rod and we predict if the rod will shoot out or in.

We use the cross product with the right hand rule to find that the force vector points outward.

Given another applet online, we played around with the same setup in a virtual environment.

We found that, since the flux is related to the area, and the change in area depends on the velocity, the change in flux is dependent on the velocity. 

Since the emf is related to the time derivative of flux, the emf is related to the velocity.

And since current is voltage over resistance, we can find the current as a function of velocity.

Now we're given the definition of inductance, we relate it to our definition of capacitance.

Since C is equal to Q / V and Q is the integral of I dt, we can use some algebra and write I in terms of C and dV/dt.

Then we define emf as the negative time derivative of flux and substitute in a formula for electric potential. Since we defined inductance as the negative emf over the time derivative of current, we can substitute again and define inductance with no time derivatives.

A sample problem where the radius, loops, and length are give, we plug in and find a value for inductance.

Above, we derive the units of inductance, we found it equal to mass times area over coulombs squared, which is defined as a Henry.

When an inductor is in a circuit, the potential difference across the inductor is proportional to the time derivative of the current. Therefore, by definition, if a sudden current change is applied to the circuit, the voltage should approach infinity. Due to that, the inductor resistant to current changes in circuits.

Here we draw the expected slope of current with an inductor in a circuit,

Here we do another Active Physics online experiment, we're given a circuit with an inductor and resistor, and the current graph.

Given another and my group's analysis of the lab, we found the scaling factors due to inductance and resistance. Then given the time constant, we can the current at given times.

Professor Mason's conclusions and derivations on the whiteboard, further concluding what we thought.