WEBVTT
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A circuit containing a resistor, a capacitor, and an inductor in series has a resonant frequency of 372 hertz.
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The resistor has a resistance of 440 ohms, and the capacitor has a capacitance of 112 millifarad.
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The peak voltage across the circuit is 28 volts.
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What is the peak current in the circuit when an alternating current in the circuit has a frequency of 372 hertz?
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Give your answer to two decimal places.
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Okay, let’s start out by sketching this circuit.
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The circuit has a resistor, a capacitor, and an inductor arranged in series.
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And because this circuit has a resonant frequency, we know that it’s an alternating current circuit.
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That is, the magnitude of the current in the circuit is not constant over time.
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Instead, it varies periodically.
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Knowing that the resistor in our circuit has a resistance of 440 ohms and that the capacitor has a capacitance of 112 millifarad, we want to solve for the peak current, that is, the maximum possible current value in the circuit.
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We want to know this value when the frequency of the circuit is 372 hertz, which we’re told is the circuit’s resonant frequency.
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Along with all this, we’re told the peak voltage in the circuit.
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It’s 28 volts.
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Let’s start solving for our peak current by writing down some of the given information.
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The resonant frequency of this circuit, we’ll call it 𝑓 sub R, is 372 hertz.
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This is the frequency at which both voltage and current can oscillate in the circuit so that the overall impedance of the circuit, which is like its resistance, is minimized.
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Along with the resonant frequency of this circuit, we’re also told its peak voltage.
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That’s 28 volts, and we’ll call it 𝑉 sub p.
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As we’ve seen, we want to solve for peak current.
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We’ll call that 𝐼 sub p.
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Knowing all this, let’s clear a bit of working space on screen.
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As we mentioned, the value of current in an alternating current circuit is not constant.
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Not only does the current magnitude vary sinusoidally for a given frequency, but even if we focus on the root mean square or rms value for current, we actually find that that value too varies with the frequency of oscillation of the circuit.
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We want to solve for this value of current here at the peak of this curve.
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And it turns out that this peak occurs at the resonant frequency of a circuit.
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Our circuit here has a resonant frequency of 372 hertz.
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And we’re told that the circuit is operating at that frequency when we want to solve for 𝐼 sub p.
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In general, current in a circuit depends on the circuit’s opposition to the flow of charge.
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In an RLC circuit, like we have here, it’s normal for both the capacitor and the inductor to contribute to that opposition.
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However, when the circuit operates at its resonant frequency, that minimizes opposition to the flow of charge.
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And 𝐼 sub p, the peak current in the circuit, is equal to the peak voltage, 𝑉 sub p, divided by the circuit resistance, 𝑅.
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Note that this equation is essentially a permutation of Ohm’s law, a law that works for direct current circuits.
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Since we’re given the peak voltage of our circuit as well as its resistance, we can substitute those values into this expression.
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And we find a result in units of amperes of 0.063 repeating.
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We’re told though to give our answer to two decimal places.
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Rounding our result, we find an answer of 0.06 amperes.
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This is the peak current value at the resonant frequency of this RLC circuit.