You have a 200-Ω resistor, a 0.400-H inductor, and a 6.00-μF capa...

Question

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. (b) What is the current amplitude?

Accepted Answer

Welcome back, everybody. We are making some observations about an L R C circuit. We're told a couple different things. We're told that it has an inductive of 0.2 Hertz. We're told that it has a resistance of four Omega's told that it has a capacitance of two micro fare adds. Now, three components are connected in series to a power supply. And we are told that the power supply has a voltage of volts and we are tasked with finding what is going to be the maximum current in Our circuit allowed. So I'm gonna move this over here. Um on top of all this, we are told as well that we have an angular frequency of 250 radiance per second. So how are we going to calculate this? Well, we have a formula for the maximum current and this is simply going to be the amplitude or the voltage of our power supply divided by the impotence of our circuit here. But what is going to be our impedance? Right, our impedance we are going to be calculating via this formula right here. It's going to be our resistance squared plus in parentheses. Here, our inductive react ints minus our capacitive reactant square. But we still need to figure out what these terms are. And luckily, we have formulas for those as well. We have that are inductive reactant is just equal to our inductive times are angular frequency and that our capacitive reactant is equal to one over R capacitance times our angular frequency. So let's calculate these values to calculate this value to calculate our maximum current allowed. All right here. So let's go ahead and start with the inductive reactant here. Following this formula. We have our conductance of .2 times our angular. Let's see here are angular frequency of 250 radiance per second. And when you plug this into your calculator, we get an inductive reactant of 50 omegas. Now, as for capacitive reactant here, we have one divided by our capacitance of two micro fare adds, but we want this unfair ads. So I'm gonna multiply this by 10 to the negative six times our angular frequency of 250 radiance per second. And this when we plug into our calculator gives us 2000 omegas Perfect. So now we are ready to calculate the impedance here. So this is gonna be Z is equal to the square root of our resistance, which is four squared plus our inductive reactant, which is 50 minus our capacitive reactant of sorry, this is going to be 2000 here. Uh And this is all going to be squared. And when we plug this into our calculator, we get 1950.4 Omega's. Now that we have found our impedance, we are ready to go ahead and find the maximum current allowed within our circuit. So I is equal to the amplitude of our power source, which is 15 volts divided by 1950.4. Omega's giving us 7.69 times 10 to the negative third amps, 47.69 mil amps, which of course corresponds to our final answer. Choice of B Thank you all so much for watching. Hope this video helped. We will see you all in the next one.

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. (b) What is the current amplitude?

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