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4bl-lab4

# 4bl-lab4 - Ross Miller 503290136 Lab 4 AC Circuits Partners...

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Ross Miller 503290136 Lab 4: AC Circuits Partners: Sam Ahn and Andre Svadjian TA: Steve Suh Session 6

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Introduction: The purpose of this experiment was to teach the students about finding inductance, capacitance and resistance in an AC circuit. We will also look at transformers and rectifiers to learn about those as well. Theory: Section 1: We began by looking at basic LRC circuits. In this section current is determined from the voltage drop across a resistor as measured by the computer, then the strength of the resistor is examined to determine the current, as following the equation I=V/R. We then look at the impedance from the equation |Z|=|V(peak)/I(peak)| to determine either the capacitance of a capacitor or the inductance of an inductor. We apply a sinusoidal voltage of 9 volts at 1 kHz to an inductor in series with a resistor and measuring the voltage drop across them in a set up like: We will use the equation |V(peak)/I(peak)|=ωL to find the value of the inductance, where I (peak) was gotten by the equation I=V/R from the known resistor of resistance 13Ohms. |Z|=| V(peak)/I(peak)|=|ωL|, this is the absolute value of the impedance of the inductor, which allows you to find the incuctance by the equation previously listed, knowing that ω=2π f this allows us to find what should be the inductance of this inductor. We will also look at how the voltage peaks and current are shifted between the inductor and the resistor because of the impedance of them both. Here is a prettier drawing of what I was trying to earlier portray: We then move on to measuring 2 inductors in parallel with each other and in series with another inductor and also in series with a resistor, in a set up that looks like: where all the inductances of the inductors are the same. The parallel capacitances could be added by the function 1/L(eq) = 1/L + 1/L => L(eq)=L/2. This system can now be viewed as 2 series inductors, one of inductance L/2 and the following of inductance L, which leads to an equivalent inductance of 3L/2.
We switch up our frequency to 5kHz and keep the voltage the same and begin to look at a single capacitor in series with the resistor we are using to again find the current from the voltage drop across the resistor by the formula I=V/R, and looking at the Voltage drop across the capacitor we will be able to determine its impedance by the formula |Z|=| V(peak)/I(peak)|, which gets us closer to finding the capacitance, so we can now use the formula |V(peak)/I(peak)|=|1/ωC|. The experiment is set up like this: or We now look at two capacitors in parallel with another in series and also in series with a resistor. Like in the previous experiment the current is determined from the voltage drop across the resistor as measured by the computer and the strength of the resistor to determine the current, as following the equation I=V/R, like in the other examples so far. We again look at the impedance from the equation |Z|=|V(peak)/I(peak)|, which follows

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4bl-lab4 - Ross Miller 503290136 Lab 4 AC Circuits Partners...

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