Saturday, April 16, 2016

4/14/16 Day 16: Inductors in Series and Parallel, Source free 1st Order Circuits, Passive RC Circuit Natural Response and Passive RL Circuit Natural Response

In class we first learned about how to find equivalent inductance when there are multiple inductors in series and in parallel. In series, the equivalent inductance is found by summing the individual inductances. In parallel, the equivalent inductance is the reciprocal of the sum of the reciprocal individual inductances. As seen, inductances are treated like resistors when summing them.

We then reviewed 1st order circuits, which are composed of source-free RL and RC circuits. A source free RC circuit consists only of resistors and a capacitor after the DC source is suddenly disconnected. In an RC circuit, the voltage across the capacitor is v=Voe^(-t/RC) and the current through the capacitor is i=(Vo/R)e^(-t/RC), where R is the Thevenin equivalent of the resistance in the circuit. The time constant in an RC circuit is tau = RC. We also found that it took about five time constants to almost completely discharge the capacitor in an RC circuit. Lastly, we did a problem on finding voltage and current in an RC circuit.

We then did a lab involving an RC circuit, titled Passive RC Circuit Natural Response. We looked at the voltage across and current through the capacitor when the DC source was disconnected.

After the lab we then learned about the second type of first order circuit, or an RL circuit. A source free RL circuit consists of resistors and an inductor. In an RL circuit, the current through the inductor is i(t)=Ioe^(-Rt/L) and the voltage across a resistor is simple the current times its resistance, or v(t) = IoRoe^(-Rt/L). Again, R in the equations is the Thevenin equivalent of the resistors in the source free RL circuit. For an RL circuit, the time constant is tau = L/R, in seconds. We also determined that it also took about five time constants to almost completely discharge the inductor in an RL circuit. We then did a simple example of an RL circuit.

Lastly, we did the lab titled  Passive RL Circuit Natural Response, which was similar to the previous lab but involved an inductor instead of a capacitor.

LECTURE:


In this problem, the equivalent inductance in the circuit was determined. Knowing that series inductors are summed and parallel inductors are reciprocally summed by the reciprocal, the equivalent inductance was found to be 15 H.


In this example, the first order DEQ of a source free RC circuit was solved using the separable integration method. The result was the voltage across the capacitor, which depends on the initial voltage, its capacitance and the Thevenin equivalent of the resistance at the ends of the capacitor. The time constant for an RC circuit is tau = RC.


In the above derivation, it was found that it would take about five time constants in a first order circuit to nearly discharge completely the capacitor or the inductor.


In this derivation, the power across the capacitor in an RC circuit was found, which is just the product of the voltage across it and the current running through the capacitor. At time equal to zero, the power across a capacitor looks very similar to the power across a resistor (Vo^2/R).


The objective of the above problem was to find the current and voltage across the capacitor in this RC circuit. It was already known that the initial voltage of the capacitor is 10 V, so all we needed to do was to simply find the time constant which is the 10 ohms times the capacitance (0.2 F), or 2 seconds. Then, the current was found by dividing the voltage by the equivalent resistance, or just simply 10 ohms.


The current through an inductor in an RL circuit was determined in the above integration of the first order DEQ. The result is the current through the inductor, which is an exponential function and depends on the initial current, the Thevenin resistance and the inductance of the inductor. The time constant for an RL circuit is tau = L/R.


In this example, the current across the inductor and the voltage of the resistor were determined in this simple RL circuit. The initial current in the circuit was provided, so all we needed to determine was the time constant, or L/R. The time constant was found to be 0.5 s, and then was used to find the current through the inductor and the voltage across the resistor, which is just the current times its resistance.


In this example, the current through the inductor was determined as a function of time in the RL circuit. First the initial current as a result of the 40 V source was determined, which was found to be 8 V. Then, using current divider the current through the inductor at t = 0 was determined to be 6 A. From there the time constant was determined in this circuit. Prior, the equivalent resistance seen by the inductor (16 ohm, 12 ohm and 4 ohm resistors) was determined to be 8 ohms. Then the time constant tau = L/R was determined and found to be 1/4 s. Lastly, using the initial current and the time constant, the equation for current was written to be 6e^-4t.


LAB:

Passive RC Circuit Natural Response:

Purpose:

The objective of this experiment was to examine the response of a simple RC circuit to a sudden removal of a DC voltage source and to an applied alternating square voltage signal. It was expected that the response method used affects the measurement of the circuit.

Prelab:


In the prelab, the expected time constant for the RC circuit when the applied square voltage signal was zero was determined. This was done by finding the Thevenin resistance then mutiplying it by the capacitance of the capacitor. The time constant was found to be about 15 ms.


The voltage of the capacitor was also determined used voltage divider. It was found to be 3.44 V for both cases, when the 5 V DC source and the square wave was added to the RC circuit.

The same was done for when the constant voltage source was completely removed from the circuit. Doing so the time constant was different, and it was found to be 48.4 ms. In this circuit the Thevenin resistance is only the 2.2 kOhm resistor. The 1 kOhm resistor is on an open circuit branch and therefore is not seen by the capacitor.

Apparatus:


The equipment used in this experiment consisted of a 22 uF capacitor,  1 kOhm and 2.2 kOhm resistors, an analog discovery, a breadboard, a laptop with Waveforms, wires, a DMM and alligator clips.

Procedure:


First of all, the circuit seen in the schematic in the prelab above was constructed, as seen in the picture above. The oscilloscope window was then used to measure the capacitor voltage, and a constant DC of 5 V was applied to the system first. Then, the 5 V source was disconnected and reconnected a few times, and the graph of the voltage across the capacitor was determined. A trigger was used in order to keep the graph in the oscilloscope obtained by removing the DC source. This action is shown in the photo below:


Data:

 The graph of the voltage across the capacitor as a result of the disconnection of the 5 V DC source is shown below:


The same was then performed with a square wave voltage. The oscilloscope window is shown below:



Data Analysis/Conclusion:

Looking at the graph of the disconnection of the DC source, the voltage of the capacitor is about 3.45 V, which is very close to our calculated 3.44 V (a 0.291% difference). In addition, the time constant was found to be 240 / 5 = 48 ms, which is also very close the calculated value of 48.4 ms. (0.826 % difference). Looking at the slope of the voltage change, the voltage does not instantaneously change, which is correlating to a capacitor's behavior.

Looking at the graph of the 5 V to 0 V square wave signal, the voltage of the capacitor is about 3.48 V, which is very close to our calculated 3.44 V (a 1.16% difference). In addition, the time constant was found to be 100 / 5 = 20 milliseconds, which is also fairly close the calculated value of 15 ms. (25 % difference). Taking into account all of the assumptions and error in measurements and readings, the values are actually very close. In addition, looking at the slope of the voltage change, the voltage does not instantaneously change, which is correlating to a capacitor's behavior.

Passive RL Circuit Natural Response:

Purpose:

The objective of this experiment was to examine the response of a simple RL circuit to a sudden removal of a DC voltage source and to an applied alternating square voltage signal. It was expected that the response method used affects the measurement of the circuit.

Prelab:


In the prelab, the expected time constant for the RL circuit when the applied square voltage signal was zero was determined. This was done by finding the Thevenin resistance then dividing the inductance of the inductor by it. The time constant was found to be about 1.45 ns.

The same was done for when the constant voltage source was completely removed from the circuit. Doing so the time constant was different, and it was found to be 0.0213 us. In this circuit the Thevenin resistance is only the 47 Ohm resistor. The 100 Ohm resistor is on an open circuit branch and therefore is not seen by the inductor.

Apparatus:

The equipment used in this experiment consisted of a 1 uH inductor,  1 kOhm , 2.2 kOhm, 47 ohm and 100 ohm resistors, an analog discovery, a breadboard, a laptop with Waveforms, wires, a DMM and alligator clips.

Procedure:


First of all, the resistances of the resistors used in this circuit were determined and recorded, as seen in the prelab. Then, the circuit seen in the schematic in the prelab above was constructed, as seen in the picture above. The oscilloscope window was then used to measure the voltages across the two resistors, and a math channel was used to calculate the current across the inductor. First, a constant DC of 5 V was applied to the system first. Then, the 5 V source was disconnected and reconnected a few times, and the graph of the voltage across the capacitor was determined. A trigger was used in order to keep the graph in the oscilloscope obtained by removing the DC source. The circuit with the 1k ohm and 2.2k ohm resistors was then used with the 0 V to 5 V square wave signal to measure the voltages and current.

Data:

 The graph of the voltage across the capacitor as a result of the disconnection of the 5 V DC source is shown below:


The same was then performed with a square wave voltage. The oscilloscope window is shown below:


Data Analysis/Conclusion:

Looking at the graph of the disconnection of the DC source, the initial current across the inductor is 3.8 mA, which correlates with our prediction, since in DC an inductor acts as a short circuit. The reason there is minute current is due to the resistance and leakage current unaccounted for in the inductor . In addition, the time constant was found to be almost instantaneous, which is also very close the calculated value of 0.0213 us. There is no way to read the graph because it is so sudden. Looking at the slope of the current change, the current does not instantaneously change, which is correlating to a inductor's behavior.

Looking at the graph of the 5 V to 0 V square wave signal, the initial current in the inductor is very small, which correlates with our prediction since it acts as a short circuit in dc. Again the reason there is some minute current is due to leakage resistance in the inductor. In addition, the time constant was found to be too small to be measured, which agrees wit our calculated value of 1.45 ns. Taking into account all of the assumptions and error in measurements and readings, the values are actually very close. In addition, looking at the slope of the current change, the current does not instantaneously change, which is correlating to a inductor's behavior.

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