Serial Rl Circuit

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Serial RL Circuit Analysis Robert Hamilton DeVry University Serial RL Circuit Analysis For week two iLab, our objectives were to analyze an AC circuit in series with the following components: 47mH inductor with internal resistance and a 470 Ω resistor. Each component will experience certain voltage drops and current alternations due to the frequency changes. The circuit was conducted at 1 kHz, 2 kHz, and 3 kHz. Given the initial conditions, we measured voltage across the components accordingly. We were also able to measure the current through the components as well. After the construction of the circuit, the simulation of the circuit followed. Overall, the iLab conducts current, voltage measurements in a series circuit. Impedance of Each Component at 1 kHz At the beginning of lab, you are given questions that require hand written calculations. First, you have to calculate the impedance of the inductor in the series circuit. Before you calculate impedance for the inductor component, you have to measure its internal resistance. In the real world, the inductor will have an internal resistance so the formula was slightly adjusted. For the 470 Ω resistor, the formula remained the same. * RL+jXL = Z1 = 74 Ω + j295 = Inductor impedance * R+j0 = Z2 = 470 Ω + j0 = Resistors impedance Total impedance and Current Once we have calculated the required information, we can apply it to solve for total impedance and total current in the series circuit. In order to solve for total impedance, you have to put the information in rectangular form first. The information above has already been converted to rectangular form so we just have to add them together to obtain total impedance. To determine the current through a certain component, we use a certain formula. This formula states that Vs/ZT will give you current. * ZT = Z1 + Z2 = (74 Ω + j295)

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