Video: Finding the Amplitude of the Oscillations in an RLC Circuit

An oscillating circuit has an inductance 𝐿 = 10 mH, a capacitance 𝐢 = 1.5 πœ‡F, and a resistance 𝑅 = 2.0 Ξ©. What is the time required for the amplitude of oscillations in this circuit to drop to half their initial value?

03:35

Video Transcript

An oscillating circuit has an inductance 𝐿 equals 10 millihenries, a capacitance 𝐢 equals 1.5 microfarads, and a resistance 𝑅 equals 2.0 ohms. What is the time required for the amplitude of oscillations in this circuit to drop to half their initial value?

We’re told these 𝑅𝐿𝐢 circuits values for inductance, capacitance, and resistance. And we want to solve for the time required for the oscillation amplitude in the circuit to drop to half its initial value. We’ll call this time 𝑑 sub one-half.

We’ll start out by recalling the mathematical equation for charge in an 𝑅𝐿𝐢 circuit like this as a function of time. As a function of time, the charge in an 𝑅𝐿𝐢 circuit is comprised of three parts.

First, there is the maximum charge that’s ever in the circuit, 𝑄 sub max. That’s multiplied by the second part, an exponential decay factor 𝑒 to the negative 𝑅𝑑 over two 𝐿. Then lastly, in part three, there is a cosine term with a driving frequency multiplied by time.

If we were to graph this function where 𝑄 of 𝑑 divided by 𝑄 max is on the vertical axis and the time 𝑑 is on the horizontal axis, then we can see the charge as a function of time oscillates up and down according to the cosine of πœ”π‘‘ times 𝑑 term. But the envelope of this function, shown here in the dotted lines, is represented by the negative exponential term 𝑒 to the negative 𝑅𝑑 over two 𝐿.

It’s that term that we want to focus on in order to answer our question of how much time needs to pass in order for 𝑄 of 𝑑 to degrade to half its initial value. That is, we’ll assume that the frequency πœ”π‘‘ is high enough that this cosine πœ”π‘‘π‘‘ term does not significantly affect the degradation of our charge as a function of 𝑑.

Neglecting that cosine term, we can now write an equation for 𝑄 as a function of 𝑑, which simply involves the maximum charge 𝑄 max and the exponential term. Now we know that when 𝑑 is equal to 𝑑 sub one-half, at that moment in time, one-half the maximum charge is equal to the maximum charge times 𝑒 to the negative 𝑅 times 𝑑 sub one-half over two 𝐿. Since 𝑄 sub max appears on both sides of the equation, we can cancel it out.

Now we want to solve the equation that remains for 𝑑 sub one-half. We can begin to do this by taking the natural logarithm of both sides, which simplifies the right-hand side to negative 𝑅 times 𝑑 sub one-half over two 𝐿. Multiplying both sides of the equation by negative two 𝐿 over 𝑅, we find this simplified equation for 𝑑 sub one-half.

We’ve been given values of inductance and resistance and can plug those in now. When we do, making sure to use units of henries for our inductance, and enter these terms on our calculator, we find that, to two significant figures, 𝑑 sub one-half is 6.9 milliseconds. That’s how long it takes the amplitude of oscillations in the circuit to degrade to one-half its maximum value.

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