Video Transcript
An alternating current circuit contains a resistor with a resistance of 25 ohms, an inductor with a 32-ohm inductive reactance, and a capacitor with a 12.8-ohm capacitive reactance. The peak voltage produced by the alternating voltage source powering the circuit is 120 volts. What is the peak current in the circuit? Give your answer to one decimal place.
Let’s begin by making a sketch of this alternating current circuit. Here, we have an AC circuit with a resistor, an inductor, and a capacitor. We’re told the resistor has a resistance of 25 ohms. And then, rather being told the inductance and capacitance of the circuit, we’re given what are called the inductive reactance and capacitive reactance. The reactance of a component in a circuit indicates how much that component opposes the flow of electric charge. Reactance is similar to, but slightly different from, resistance. We see though that reactance and resistance are similar enough that they share the same units, units of ohms. The symbol representing reactance is a capital 𝑋. For the inductor’s reactance, we’ll write 𝑋 sub 𝐿, and for the capacitive reactance, 𝑋 sub 𝐶.
We’re told what these two values are, again in units of ohms. Now, since this is an alternating current or AC circuit, this means that voltage, like current, varies in the circuit. It follows a sinusoidal pattern. We’re told that the very highest value of this voltage, we’ll call it 𝑉 sub P, is 120 volts. Knowing all this, the first part of our question asks, “What is the peak current in the circuit?” So we know the peak voltage, and we want to solve for the peak current. Peak current, we’ll represent it as 𝐼 sub P, is related to peak voltage in a way that’s similar to Ohm’s law.
Recall that Ohm’s law states that the potential difference 𝑉 across a circuit equals the current in the circuit 𝐼 multiplied by the circuit resistance 𝑅. Dividing both sides by 𝑅, we have 𝐼 equals 𝑉 over 𝑅. The equation giving 𝐼 sub P is like this. It’s equal to the peak voltage divided by something called the impedance of the circuit. In an AC circuit, impedance is a quantity that combines all quantities that oppose the flow of charge. This includes all resistances and reactances. To find the impedance 𝑍 of our circuit, we might think we simply need to add together these values given in units of ohms. But it’s a little more complicated than that.
Let’s clear some space on screen, and we’ll use this space to recall the mathematical equation for impedance 𝑍. Notice that it involves resistance 𝑅, as well as the difference between inductive reactance and capacitive reactance. Thankfully, we have all of that information for our given circuit. Substituting these values in, we have the impedance of our circuit equals the square root of 25 ohms squared plus the quantity 32 ohms minus 12.8 ohms squared. Now, we don’t need to calculate 𝑍 quite yet because, remember, we’re looking for the peak current in the circuit. That quantity equals the impedance 𝑍 divided into the peak voltage. Substituting in 120 volts for 𝑉 sub P, now we’re ready to calculate this peak current. Rounded to one decimal place, we find it’s 3.8 amperes.
So, in other words, if we saw how 𝐼 varies with time in this AC circuit, the peak value on that curve would be approximately 3.8 amperes. That’s our answer to part one of this question. Let’s move on now to part two.
What is the root-mean-square current in the circuit? Give your answer to one decimal place.
Let’s start by thinking about this idea of root-mean-square current. We know that the current in an alternating current circuit varies with time. So if someone said, “what is the current in this circuit?” we couldn’t give just one answer. That’s where the root-mean-square current comes in. This is a mathematical measure of current that varies sinusoidally so that we can report one single representative current value. And that value stays constant over time.
To see how this works, imagine that we have this current in a circuit over time. If we wanted one representative current value for this alternating current, we could report the average current value. But then that would just be zero. That’s not a great representation of this alternating current. The idea behind root-mean-square current is that, first, we square all of the individual current values in our curve. If we did that, we would get a curve that looks something like this pink one. Notice that none of the values in this curve are negative. They’re all either positive or zero.
Next, what we do is we find the mean or average value of all these points in our pink curve. Let’s say that that average value is somewhere around here. Lastly, what we do is take the square root of this average. That would drop the line down to be maybe here, say, where the blue line is. This value then is what we call the root-mean-square current in the circuit. Notice that it’s not zero, even though zero is the average value of the current.
Now, it turns out that the peak current in a circuit and the root-mean-square current are related through this equation. The root-mean-square current equals the peak current divided by the square root of two. That’s actually all we need to do to solve for the root-mean-square current from the peak current. And since we’ve already solved for the peak current in part one, we can use that value to solve for 𝐼 sub rms. Rounding our answer to one decimal place, we get a result of 2.7 amperes. This is the root-mean-square current in this alternating current circuit.
