A series RLC circuit consists of a 75 Ω resistor, a 0.12 H ind...

Question

A series RLC circuit consists of a 75 Ω resistor, a 0.12 H inductor, and a 30 μF capacitor. It is attached to a 120 V/60 Hz power line. What is the peak current I?

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A speaker in a home audio system represented as a series RLC circuit with a 30 ohms resistor, 0.4 Henry's inductor and 50 micro Ferras capacitor is driven by an 80 Hertz 60 volts R MS amplifier determine the peak current flowing through the speaker. So that is our end goal. Our end goal is we're trying to find what the peak current value will be that flows through the speaker. Awesome. We're also given some multiple choice answers for the peak current and they're all in the same units of AMP here. So let's read them off to see what our final answer might be. A is 0.82 B is 0.52 C is 0.21 and D is 0.61. Awesome. So first off, let us note that the speaker is represented as an RLC circuit which stands for resistor inductor capacitor. And this RLC circuit is in a series. The current in the circuit depends on the frequency of the input and the values of the RL and C elements. So now let's write down all of our known variables. So we don't get confused. We're given our frequency, which we're gonna represent it with F is equal to 80 Hertz. We're also given the resistance, we're gonna represent it by capital R and the resistance is 30 ohms. We're also given the inductance, which we're gonna represent it by capital L is equal to 0.4 Henry. We're also given the capacitance, which you're gonna represent with a capital C which is equal to 50 micro farads, which is also equal to 50 multiplied by 10 to the power of negative six fads. We're also given the R MS voltage which the, which R MS stands for root means square. So the root means square voltage or R MS voltage for short, which we're gonna denote as V subscript R MS is equal to 60 volts. Awesome. So at this point, we need to calculate the reactance of the inductor and the capacitor. So let's denote our inductive reactant as capital X subscript capital L. And let's also represent our capacitance reactants as capital X subscript C. And let's write equations for both of those. So we need to recall and use the equations for the inductive reactants and the capacitance reactants. Fantastic. So let's start with inductive reactants. So XL is equal to two multiplied by pi multiplied by the frequency multiplied by the inductance. So when we plug in our known variables, we get two multiplied by pi multiplied by 80 hurts multiplied by 0.4. He when we plug that into a calculator, we will get 201.06 ohms. Awesome. So now recalling the equation for the capacitance reactants, so XC is equal to one divided by two multiplied by pi multiplied by the frequency multiplied by the capacitance. So like above, let's plug in our known variables. So one divided by two multiplied by pi multiplied by 80 hurts multiplied by the capacitance which was 15, multiplied by 10 to the power of negative six fair adds. So when we plug that into a calculator, we will get 39.788 which rounds to 39.79 ohms. Awesome. So now we need to use these values that we just found to help us solve for the total impedance of the circuit. So let's represent the total impedance as Z. So the total impedance is equal to the square root. So when we recall this equation, we remember that the total impedance is equal to the resistance squared plus the inductive reactants subtracted from the capacitance reactants. And this is squared. So now we can plug in our known variables to solve for Z which is the total impedance. So we know that the resistance is 30 ohms squared plus XL which is 201.06 ohms minus 39.79 ohms and this is all squared. Awesome. So when we plug that into a calculator, we will get that. The total impedance is equal to 164.0366 which when we round to two decimal places, we will get 164.04 ohms. Awesome. We're moving right along. So now we need to solve for the R MS current and use that value to solve for the peak current value that we are solving for which is our end goal. Awesome. So, so for the RMS current, let's denote it as capital I subscript RMS. So to solve for the RMS current, we will recall that the R MS current is equal to the root means square voltage, the R MS voltage divided by the total impedance. So let's plug in our known variables and solve. So our R MS voltage was 60 volts divided by the total impedance which we discovered that value to be 164.04 arms. So when we plug that into a calculator, we will get 0.3 66 and piers. So we will recall that the peak current. So let's call it capital I subscript P for the peak current is equal to the square root of two multiplied by our R MS current, which is 0.366 years. So when we plug that into a calculator, we will get 0.5176 which will be round to two decimal places. Like all the multiple choice answers are we will get 0.52 and piers. So this means that the peak current is approximately 0.52 amps or amp here. So when you round the two decimal places, you will be 0.52. Awesome. So let's box our answer in green and celebrate the fact that we solved for our final answer. Hooray. So looking at our multiple choice answers, that means the correct answer has to be the letter B 0.5 pi Thank you so much for watching. Hopefully that helped and I can't wait to see the next video. Bye.

A series RLC circuit consists of a 75 Ω resistor, a 0.12 H inductor, and a 30 μF capacitor. It is attached to a 120 V/60 Hz power line. What is the peak current I?

Key Concepts

Impedance in RLC Circuits

Ohm's Law for AC Circuits

Resonance in RLC Circuits

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