You have a 200-Ω resistor, a 0.400-H inductor, and a 6.00-μF capacitor. They are connected to form an L-R-C series circuit. (a) What is the impedance of the circuit?

Video transcript

Welcome back everybody. We are making observations about a circuit and we are told that it has an induct er with conductance of 125 mil. It hurts a resistor with a resistance of Omegas and a capacitor with capacitance of 400 micro fare adds. Now, they are all connected in Series two A time varying power source that has a peak value of 12V and an angular frequency of 450 radiance per second. Now, we are tasked with finding what the impedance is of our LRC circuit. Here, we have a formula for this impedance is given by our resistance squared plus our inductive reactant minus our capacitive reactant squared. But we need these two values as well. And luckily, we have formulas for those as well. Inductive reactant is given simply by conductance times our angular frequency. And capacitive reactant is given by one over R capacitance times our angular frequency. Let's go and calculate these terms and go ahead and then find our impedance here. All right. So let's start with the inductive react ints first plugging in our values. We get 125 mila hurts but we want this and hurt. So I'm gonna multiply this by 10 to the negative three times our angular frequency of 450 giving us 56.25 Omega's. Now as for our capacitive reactant, we have one over our capacitance of 400 micro fare adds, but we wanted a fair at. So I'm gonna multiply this by 10 to the negative six times our angular frequency of 450 radiance per second giving us 5. Omega's here. Right now. We are ready to go ahead and calculate our Z value here. So this is going to be the square root of squared plus 56.25 minus 5.55 squared, which when we plug into our calculator gives us a final answer of 230. Omegas which corresponds to our final answer. Choice of C Thank you all so much for watching. Hope this video helped. We will see you well in the next one.