Video: Eg17S1-Physics-Q23

When a capacitor is connected in series to the circuit shown, the reading of the hot-wire ammeter does not change. What is the ratio of the capacitive reactance of the capacitor to the inductive reactance of the coil?

05:44

Video Transcript

When a capacitor is connected in series to the circuit shown, the reading of the hot-wire ammeter does not change. What is the ratio of the capacitive reactance of the capacitor to the inductive reactance of the coil?

Alright, so in this circuit, we first of all got an AC source. And then connected in series with this AC source, we’ve got a resistor with resistance 𝑅, a coil with inductance 𝐿, and an ammeter. We’ve been told that this ammeter is a hot-wire ammeter and that when we connect a capacitor in series with this circuit. So, for example, if we connect a capacitor here and say that it has a capacitance 𝐢, then we’re told that the reading on the hot-wire ammeter does not change.

Now the hot-wire ammeter is measuring the current through the entire circuit because, remember, this is a series circuit. Therefore, the current through all of the components is exactly the same. And hence, what we’re told is that when the capacitor is connected in series with this circuit, the current through the circuit does not change because the reading on the hot-wire ammeter does not change.

In other words then, we can say that the current through the circuit before we connect the capacitor β€” that’s 𝐼 subscript before β€” is equal to the current through the circuit after we connect the capacitor. And in this question, we’ve been asked to find the ratio of the capacitive reactance of the capacitor to the inductive reactance of the coil. In other words, we’ve been asked to find the ratio 𝑋 subscript 𝐢, which is the capacitive reactance of the capacitor, to 𝑋 𝐿. That’s the inductive reactance of the coil. Where, of course this is the capacitor and this is the coil.

Now in order to find this ratio, we first need to recall that the impedance of an RLC circuit. That’s the impedance of a circuit containing some sort of resistance, some sort of inductance, and some sort of capacitance. Is given by the square root of the total resistance of the circuit squared plus the square of the inductive reactance minus the capacitive reactance. And this is true for a general RLC circuit.

So let’s think about what the impedance of the circuit was before we connected the capacitor. Let’s call this impedance 𝑍 subscript before. And we can see that it ends up being the square root of 𝑅 squared, because the resistor is already in the circuit, plus 𝑋 𝐿 squared. And the reason for this is that, before we connected the capacitor to the circuit, there was no capacitor in the circuit at all. Therefore, there is no capacitive reactance.

However, after we connect the capacitor, we can say that the impedance of the circuit, 𝑍 subscript after, is given by the square root of 𝑅 squared plus 𝑋 𝐿 minus 𝑋 𝐢 squared, just like the general equation. At which point we have expressions for the impedance of the circuit before we connected the capacitor and the impedance after we connected the capacitor. So why is this important in the first place?

Well, it’s important because we’ve been told that the current in the circuit does not change when we connect the capacitor. And as well as this, before we connect the capacitor or after we connect the capacitor, we’re using the same AC source. Therefore, the potential difference in the circuit does not change either. We can state this as the potential difference across the circuit before is equal to the potential difference across the circuit after we connect the capacitor.

At which point we can remember the version of Ohm’s law which applies for circuits which contain impedances and capacitances as well as resistances. In this situation, Ohm’s law tells us that the potential difference across the circuit is equal to the current through the circuit multiplied by the impedance of the circuit. And then we can use this version of Ohm’s law to give us two separate equations for before and after we connect the capacitor.

In other words, the total potential difference across the circuit before we connect the capacitor is equal to the current through the circuit before we connect it multiplied by the impedance of the circuit before we connect it. And the same is true for after. We can use 𝑉 after, 𝐼 after, and 𝑍 after. But then, at this point, we can see that we’ve said that 𝑉 before is equal to 𝑉 after. That’s this equation here. And 𝐼 before is equal to 𝐼 after as well. That’s this equation here. So if 𝑉 before is equal to 𝑉 after and 𝐼 before is equal to 𝐼 after, then the only way that both equations can be true is if 𝑍 before is equal to 𝑍 after as well.

We can see this by replacing 𝑉 before by 𝑉 after, for example, because they’re the same thing. And we can also replace 𝐼 before by 𝐼 after, at which point the top equation tells us 𝑉 after is equal to 𝐼 after multiplied by 𝑍 before. But it’s also equal to 𝐼 after multiplied by 𝑍 after. And so 𝑍 before must be equal to 𝑍 after. So we just found out that 𝑍 before is equal to 𝑍 after, which means that the square root of 𝑅 squared plus 𝑋 𝐿 squared is equal to the square root of 𝑅 squared plus 𝑋 𝐿 minus 𝑋 𝐢 squared. We’ve basically equated this with this.

So let’s give ourselves a little bit more space and write the same thing up here and then work through this. Let’s start by squaring both sides of the equation, which means that the squareds cancel with the square roots on both sides. This leaves us with 𝑅 squared plus 𝑋 𝐿 squared is equal to 𝑅 squared plus 𝑋 𝐿 minus 𝑋 𝐢 whole squared. At which point we can see that we’ve got an 𝑅 squared on both sides. So let’s subtract 𝑅 squared from both sides of the equation. This way, 𝑅 squared cancels on both sides.

So now we’re left with 𝑋 𝐿 squared is equal to 𝑋 𝐿 minus 𝑋 𝐢 whole squared, which means that now we should expand this bracket. Doing so leaves us with 𝑋 𝐿 squared on the left and 𝑋 𝐿 squared minus two 𝑋 𝐿 𝑋 𝐢 plus 𝑋 𝐢 squared on the right. And now we see that we’ve got an 𝑋 𝐿 squared on both sides. So we subtract 𝑋 𝐿 squared from each side. What we’re left with then is zero on the left-hand side and negative two 𝑋 𝐿 𝑋 𝐢 plus 𝑋 𝐢 squared on the right.

Now let’s rearrange by adding two 𝑋 𝐿 𝑋 𝐢 to both sides, which means that it cancels on the right-hand side. And then we can divide both sides of the equation by 𝑋 𝐢. So that one factor cancels on the left and one factor cancels on the right. This leaves us with two 𝑋 𝐿 is equal to 𝑋 𝐢. In other words, the capacitive reactance of the capacitor is twice as large as the inductive reactance of the coil.

Now in the question, we’ve been asked to find the ratio 𝑋 𝐢 to 𝑋 𝐿. That’s this ratio here. And because 𝑋 𝐢 is twice as large as 𝑋 𝐿, we can say that the ratio is two to one, at which point we’ve arrived at the final answer. The ratio of the capacitive reactance of the capacitor to the inductive reactance of the coil is two to one.

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