Video Transcript
An RLC circuit contains a capacitor, an inductor, and an ohmic resistor connected in
series. The ohmic resistor has a resistance of 60 Ω. When the frequency of the ac source is 𝑓 naught, the maximum current possible, 𝐼
sub 𝑚, is present in the circuit. And when the frequency of the ac source is increased to two 𝑓 naught, the current
present in the circuit decreases to 𝐼 sub 𝑚 over three. Calculate the reactance of the inductor when the ac source frequency is 𝑓
naught.
Here we have an RLC series circuit, and we want to solve for its inductive
reactance. Let’s start by recalling a few formulas relevant to a circuit like this. Answering this question involves quite a bit of algebra, so it’ll be good to keep
these formulas and all our work organized.
First, let’s take a look at the general formula for the impedance, 𝑍, of a circuit,
where 𝑅 is the resistance of the circuit, 𝑋 sub 𝐿 is its inductive reactance, and
𝑋 sub 𝐶 is its capacitive reactance. It’s good to note that the only constant term in this expression is resistance. Recall that the reactance values, and thus the overall value of inductance, vary with
the frequency of the ac source.
Now, this question is asking us to find 𝑋 𝐿 when the ac source frequency is 𝑓
naught. And we were told that at this frequency, the current in the circuit is at a maximum,
𝐼 𝑚. We were also told that when the frequency is two times that value, so two 𝑓 naught,
the current decreases by a factor of three.
There are a few more pieces of information we should briefly review before we get to
work. Let’s next recall that when the ac source is at the resonant frequency, the current
in the circuit is maximized. So in this case, the resonant frequency is 𝑓 naught. At the resonant frequency, the impedance of the circuit is minimized and is simply
equal to 𝑅, the resistance of the circuit. Looking back at the impedance formula in this case, this entire term, the inductive
reactance minus the capacitive reactance, must equal zero. Last, let’s recall the formula for each of these reactance values. The inductive reactance, which we want to solve for, equals two 𝜋𝑓 times the
inductance, 𝐿, of the circuit. And the capacitive reactance equals one over two 𝜋𝑓 times 𝐶, the capacitance of
the circuit.
Okay, now that we’ve refreshed our memories, let’s get to work solving for 𝑋 𝐿,
specifically its value when the ac source is at the resonant frequency. We’ve already established that at the resonant frequency, the inductive reactance
minus the capacitive reactance equals zero. So we can say that these two reactance values are equal. Let’s expand upon this by substituting in the expressions for each value, noting that
𝑓 equals 𝑓 naught here. So we have that two 𝜋𝑓 naught 𝐿 equals one over two 𝜋𝑓 naught 𝐶. Now, we know that this circuit contains an inductor and a capacitor, but we don’t
actually know a value for either 𝐿 or 𝐶. So we should rearrange this expression to solve for one of them.
In this case, let’s just choose to make capacitance the subject of the equation. Doing a little algebra, we get that 𝐶 equals one over four 𝜋 squared 𝑓 naught
squared times 𝐿. This expression will be really useful to us since the capacitance of the circuit is
constant, even though the capacitive reactance isn’t. Let’s leave this here for now and move on to use the formula for the overall
inductance of the circuit when the ac source is at twice the resonant frequency. We’ll substitute in the formula for 𝑋 𝐿. But first, let’s rewrite the expression 𝑋 𝐶 using the value of capacitance that we
just solved for.
After doing a little simplifying, we get that 𝑋 𝐶 equals 𝜋 times 𝑓 naught times
𝐿. So this is what we’ll substitute into the formula for inductance. We also need to substitute in the value of 𝑋 𝐿 at twice the resonant frequency,
which equals four 𝜋𝑓 naught times 𝐿. After doing so, our expression for the impedance when the ac source is at twice the
resonant frequency can be written like this. Then, distributing this power of two gives 𝑅 squared plus nine 𝜋 squared 𝑓 naught
squared 𝐿 squared all to the power of one-half equals 𝑍.
Hang in there — we’re getting closer! We have this big expression for impedance, and we need a way to relate it to some
values we already know. Luckily, we know that 𝑍 is equal to 𝑅 when the circuit is at the resonant
frequency. This will be really helpful to us, so let’s take a closer look. We can apply Ohm’s law, which states that resistance equals potential difference
divided by current. And we know that at the resonant frequency, the current is maximized and equals 𝐼
sub 𝑚.
But what about when the ac source is at twice the resonant frequency? Well, we know that the resistance of the circuit doesn’t change but that the current
decreases by a factor of three. Assuming that the applied potential difference is constant, then if current decreases
by a factor of three, the impedance must increase by a factor of three. Therefore, at twice the resonant frequency, 𝑍 equals three times 𝑅. So at this frequency, we know an expression for impedance in terms of the resistance,
which is a known value.
So, setting this all equal to three 𝑅, let’s now work toward solving for the
inductance 𝐿. Then, once we have that, we’ll be able to plug it into the formula for 𝑋 𝐿 and
we’ll have our final answer. First, we can square both sides of the equation. Then, we should work to get 𝐿 by itself on one side of the equals sign by
subtracting 𝑅 squared from both sides. This gives us eight 𝑅 squared equals nine 𝜋 squared 𝑓 naught squared 𝐿
squared. So let’s next divide both sides by nine 𝜋 squared 𝑓 naught squared. And those terms will cancel out of the right-hand side, leaving behind only 𝐿
squared. So now, taking the square root of both sides and simplifying, we find that 𝐿 equals
two root two 𝑅 divided by three 𝜋𝑓 naught. And we know that this value is constant.
All that’s left to do now is substitute the inductance into the formula for inductive
reactance. Remember that we want to know the reactance of the inductor when the ac source is at
the resonant frequency. So we’ll use 𝑓 naught for the value of frequency. Substituting the inductance into this formula, notice that 𝜋 and 𝑓 naught cancel
out of the numerator and denominator. So we’re left with 𝑋 𝐿 equals one-third times four root two times 𝑅. We were told that the resistance of the circuit equals 60 ohms. So substituting this in and simplifying, we reach our final value of the inductive
reactance.
Thus, we’ve found that when the ac source is at the resonant frequency, the inductive
reactance is equal to 80 times root two ohms.
