Lab 23: RL Circuits

We were give a voltage vs time graph, and asked to graph current vs time in an inductor. First we  said that ε = flux / I, we solve for current and then take the derivative so that we could see the relation between I and ε. We realized that they have inverse relationship, so the graph should also be inverse with respect to time. Then we were given a resistor, and we determined the corresponding magnitude of the resistor by looking at the colors it has. In this case, it was brown, black, and brown. We used the resistor-color table for this. 

 Then we were given an inductor, with the given variables we used the equation L = (μ0AN^2)/l, to calculate its inductance. Then we used the formula Rcoil = pL/A to calculate the resistance of the copper coil, for the area we used the cut or cross sectional area. Then we calculated the time constant with the inductance previously calculated and the 150ohms resistor.

We start working on the lab manual. This lab consisted on a circuit using a oscilloscope, function generator, resistor and inductor.

On the function generator, we set the frequency at 40kHz and the voltage at 2V, and we used the square wave form to graph whats on the picture.

 We used an inductor with 55 turns, and measure its theoretical inductance, which was 190μH. From the given graph, we measure the half-time t1/2  of the decay of the induced emf. In order to get a precise value, we adjusted the horizontal and vertical scales so that a single decay fills the oscilloscope display. We found that the half-time is 1.5μs. Then using using the formulas τ = L/R and τ = t1/2 /ln2, we equate them in order to find the experimental inductance, which turn out to be 324μH, then using this inductance, we calculated the corresponding turns, which was N = 72 turns.

To conclude the class, we worked on a RL Circuit problem. We were given the values of inductor, 2 resistors, and voltage supply. To find the time constant when the switch is first closed, we used the formula τ = L/R, where L = 35mH and for R we used the 120ohms resistor since the switch is closed. Then as the switch opens again, we were asked to find the current. Since the resistors are in parallel, the voltage across each resistor should be 45V, so we used Ohms law I = V/R to calculate the current in the 730 resistor. For the 120 resistor we cannot only use Ohms law since it is in series with an inductor, so we used I = Imax(1-e^(-t/τ )), where Imax can be found using Ohms law. The voltage drop for both resistor can be calculated by using V = IR. For part d, we used the same formula as when we calculated the current in the 120 resistor, but we solved for the time using V = 34V. Lastly, the energy dissipated can be calculated by using E = (1/2)LI^2.