Impedance Lab

Objective:
We covered phasor relationships for circuit elements. Based on the acronym "ELI the ICE man", we are able to see what value leads what. For instance, ELI means that for an inductor, the voltage leads the current. For ICE, in terms of dealing with an inductor, the current leads the voltage. Finally, for the resistor, the voltage is in phase with the current. The following are the voltage current relations for three passive elements: V=IR for resistor, V = jധLI for an inductor, V = I/jധC for a capacitor. Theses equations can then be written in terms of a ratio of the phasor voltage to current in terms of Z which represents the impedance. the impedance may be represented in rectangular form, Z= R+jX where R is the resistance and X is the impedance of the element being dealt with. We can then apply this to Kirchhoff's Laws in the Frequency Domain.

Group Practice:
1. The problem below shows a voltage source of 10cosωt connected in series with a resistor of 5Ω and a capacitor of .1F. We write the impedance and solve for it as seen below. We also solve for the current by using I=V/Z where the Z is changed to polar form. It is easier to divide using polar form. We now solve for the voltage across the capacitor using equations V = IZc. The final results will be in the time domain.

Figure 1. Solving for the current and the voltage across the capacitor.

2. The same situation is modeled as in figure 1. However, there is a current source and a capacitor to deal with. We solve for the impedance

Figure 2. Solving for voltage across the capacitor involving a current source.

3. The problem below shows a circuit with an inductor parallel to a capacitor. In order to solve this, we need to find the impedance of the elements and find their equivalent. We can use a voltage divider in order to find the voltage across the 5H inductor. We keep in mind that we will need to change between rectangular to polar form. 

Figure 3. Solving for the voltage across the inductor in terms of the time domain.

Impedance Lab Procedures and Results:

1. For our pre lab, we are required to calculate the impedances of the resistor, inductor and capacitance as seen in figure 4. We will also find the current for each component in terms of it magnitude and the phase angle as seen in figure 5. 

Figure 4. Finding the impedance of each circuit.

Figure 5. Solving for the current in terms of magnitude and phase angle.

2. We construct the circuits from figure 4 and measure the voltage across the 47Ω resistor for each component connected in series, the current and the voltage across each components as seen in figure 4.  The measurements for the components are as follows: Resistance = 46.9士 .05Ω, R = 98.8士.05Ω, C = .092uf士.005, L = 1mH. 

3.  A 47ohm resistor will be connected in series with a a resistor R = 98.8ohms. Then the resistor will be changed to a capacitor C=.092uf with the 47 ohm resistor still connected in series. Finally, the capacitor will changed out to an inductor L = 1mH with the 47 ohm resistor still in series. The oscilloscope measures the voltage across the component mentioned, the current which is mathematically written as Voltage across the 47ohm resistor divided by the resistor value, I=Vr/R. The measurements can be seen below. The voltage input for all experiments are v(t) = 2cosωt. we also acquire our frequency by ω=2πf where f is 1kHz,5KHz and 10KHz

Figure 6. 1KHz with resistor

Figure 7. 5KHz with resistor

Figure 8. 10KHz with resistor

Figure 9. 1KHz with inductor

Figure 10. 5KHz with inductor

Figure 11. 10 KHz with inductor

Figure 12. 1KHz with capacitor

Figure 13. 5KHz with capacitor

Figure 14. 10KHz with capacitor

4. Out theoretical results are compared to experimental results as seen in figure 15.

Figure 15. Theoretical and experimental results

Learning Outcome: 

The lab requires finding the impedance Z of three circuits that contain a resistor Resistance = 46.9士 .05Ω  to another resistor R = 98.8士.05Ω, resistor Resistance = 46.9士 .05Ω, connected to an inductor  L = 1mH, and resistor Resistance = 46.9士 .05Ω, connected to a capacitor C = .092uf士.005. We are told to input a sinusoidal voltage of v(t)= 2cos(ωt) where 2 is the amplitude and ω is the frequency. When calculating the theoretical impedance, we must point out that our given freqencies of 1kHz, 5kHz, and 10kHz must be changed to rad/sec, so the equation ω=2πf will be used. Though, for the experiment, the frequency in hertz is used. Calculations for impedance for each set up can be seen in figure 15, as well as the current in the form of time domain. We are able to compare our theoretical and experimental amplitudes for the current which shows a percent differene of less than 6% for all sets up meantioned above. We can see based on out theoretical calculations that for the resistor connected in series with another resistor that voltage is in phase with current not matter whhat the freqnency. However, for the circuits that are connected with the resistor in series with the inductor or capacitor are different. For the inductor, the voltage leads the current and for the capacitor, the current lead the current. We ere unable to find the phase difference between the current and the voltages.