Video Transcript
A circuit containing a capacitor and an inductor in series has a resonant frequency of 155 kilohertz. The capacitor in the circuit has a capacitance of 215 microfarads. What is the inductive reactance of the circuit? Give your answer in scientific notation to two decimal places.
Here, we want to solve for what’s called the inductive reactance of our circuit. That circuit looks something like this. We have an alternating current and a capacitor and inductor in series. The capacitance of the capacitor, we’ll call it 𝐶, is 215 microfarads. And the resonant frequency of our circuit, we’ll call it 𝑓 sub R, is 155 kilohertz.
Knowing all this, we want to solve not for the inductance of our circuit but the inductive reactance. The inductive reactance is a measure of just how much this inductor opposes the flow of charge in our circuit. Written as an equation, inductive reactance looks like this: capital 𝑋 sub 𝐿, where the capital 𝑋 tells us it’s reactance and the 𝐿 tells us we’re talking about inductive reactance. Notice that inductive reactance is proportional to the frequency of oscillation of a circuit and its inductance 𝐿.
In our circuit, we know it’s frequency of oscillation, the resonant frequency, but we don’t know the inductance of its inductor. We can’t yet solve for inductive reactance then, but we can solve for it by an indirect route. The resident frequency of our circuit, which we know, depends on the capacitance and the inductance of the circuit. Mathematically, the resonant frequency 𝑓 sub R of a circuit equals one over two 𝜋 times the square root of 𝐿 times 𝐶, the inductance and the capacitance in that circuit. So, we can use this equation for resonant frequency along with our known values to solve for the inductance for our circuit 𝐿 and then use that value to solve for the inductive reactance.
Let’s start by clearing some space and rearranging this equation for resonant frequency so that the inductance 𝐿 is the subject. We’ll start by multiplying both sides by the square root of 𝐿 over 𝑓 sub R. This cancels the resonant frequency on the left and the square root of 𝐿 on the right. That gives us this equation. And if we square both sides, we find that the inductance 𝐿 equals one over four 𝜋 squared times 𝑓 sub 𝑅 squared times 𝐶.
Now, remember that it’s not the inductance we want to solve for exactly but the inductive reactance of our circuit. Here’s what we’ll do then. We’ll take the equation for inductive reactance, using as our frequency the resonant frequency, and replace the inductance 𝐿 with this expression here. When we do that, assuming that our circuit is at the resonant frequency 𝑓 sub R, we see some cancellation occurs. In numerator and denominator, one factor of two 𝜋 cancels out and then also one factor of 𝑓 sub R, the resonant frequency, cancels. That leaves us with this expression for the inductive reactance.
Since we know the resonant frequency of our circuit as well as its capacitance 𝐶, we can substitute those values into our expression for 𝑋 sub 𝐿. 155 times 10 to the third hertz for 𝑓 sub R and 215 times 10 to the negative sixth farads for 𝐶. Calculating this result, we get an answer, to two decimal places, of 4.78 times 10 to the negative third ohms. This is the inductive reactance of our circuit, and note that it’s in units of resistance, ohms. This confirms to us that reactance is indeed a measure of the opposition to charge flow.
