Video: Physics Past Exam • 2017/2018 • Pack 1 • Question 22

Physics Past Exam • 2017/2018 • Pack 1 • Question 22


Video Transcript

An inductor, a capacitor, a resistor, and an ammeter are connected in series with an AC source. The circuit is at resonance. If a soft iron bar is placed inside the inductor coil, the reading of the ammeter will a) increase, b) decrease, c) remain the same, d) equal zero.

Let’s start off by drawing a diagram of this circuit. In this circuit, we’re told we have an AC power supply, an inductor, a capacitor, a resistor, as well as an ammeter to measure current. At the outset, our inductor has nothing in its coils but air. And we’re told that, under these conditions, the circuit is at resonance.

Keeping that in mind, we want to know if we take a soft iron bar and place it within the windings of this inductor coil what will happen to the reading of the ammeter in the circuit, that is, what will happen to the circuit current.

To answer this question, let’s begin by thinking about what the current is like in a circuit like we have here. Because we have an AC source, we know that the current value will change in time. It will move in a sinusoidal wave. If we plotted a segment of that wave, we would see that the current is constantly changing in time. But that’s not what our answer choices refer to.

What we want to look out for is whether inserting this soft iron bar into our inductor coil will cause the current to shift overall up or down or stay the same or continue to be zero. In other words, if we kept an eye out on the maximum current in this circuit and saw how that value changed when we do and do not have this iron bar inside our inductor coil, that would help us answer our question.

We’ll refer to this maximum current value as 𝐼 sub max. And as an equation, 𝐼 sub max is equal to the maximum voltage in the circuit divided by its overall impedance, 𝑍. If we bring this equation front and center, we see that what we’re trying to find out is whether 𝐼 sub max goes up and down, stays the same, or remains zero with the insertion of this iron bar.

To figure out just how this current does change, we’ll want to look at how the terms that it’s equal to might change. Now 𝑉 sub max, the maximum potential difference created by our source, won’t be affected by whether or not our inductor coil has an iron bar in it. 𝑉 sub max will be the same value regardless.

However, the impedance, 𝑍, of our circuit will be affected by this change. Let’s see how that is. The impedance of an RLC circuit, like we have here, is equal to the square root of the resistance in the circuit squared plus the quantity the inductive reactance, 𝑋 sub 𝑙, minus the capacitive reactance, 𝑋 sub 𝑐, all squared.

In our circuit, because we do have a resistor, a capacitor, and an inductor, we can assume that all of these terms by themselves are nonzero and that they play into our impedance, 𝑍. But let’s think about our circuit before we insert our soft iron bar.

At that point, we’re told the circuit is at resonance. That has a very specific meaning for the impedance of the circuit. At resonance, the inductive reactance is equal to the capacitive reactance, which means that this term in our square root sign is equal to zero.

We could say then that 𝑍 sub 𝑟, the impedance of our circuit at resonance, is equal simply to the resistance 𝑅 of the circuit. As we consider the expression for impedance and this resulting resonant impedance, we see that, at resonance, the impedance of our circuit is as small as it can ever be. Any change in the inductive or the capacitive reactance off of resonance will increase the impedance of our circuit.

When we are at resonance and our impedance, 𝑍, is equal simply to 𝑅, that means the denominator of this fraction is as small as it will ever be and therefore the fraction is as large as it will ever be. Here then is how we can summarize the current in our circuit when we’re at resonance, before the iron bar is inserted.

We can write that when there’s no iron bar in our inductor coil, our maximum current is equal to 𝑉 sub max divided by 𝑅, the resistance of the circuit. We’re now ready to consider what will happen when we insert this iron bar into the coil. Here’s the question we want to ask when we consider this bar in the coil. Does the insertion of the bar affect the impedance, 𝑍, of our circuit? If it does, by a relationship for 𝐼 sub max, we know that that effect on impedance will have an effect on current.

To see if the iron bar will affect our impedance, let’s consider what changes it might have on the terms in 𝑍. Will inserting the iron bar affect the resistance in our circuit? No, because that value comes from the resistor, which is independent of the bar. Will the iron bar have an effect on the capacitive reactance of our circuit? No, because that depends on the capacitor. But will the iron bar have an effect on the inductive reactance of our circuit? Well, just what is the inductive reactance?

As an equation, 𝑋 sub 𝑙 is equal to two 𝜋 times the frequency of oscillation of the circuit multiplied by the self-induction coefficient of the inductor. And based on this equation, we can recall the mathematical relationship for that self-induction coefficient 𝐿. 𝐿 is equal to the number of turns in a coil, 𝑁, squared multiplied by its cross-sectional area divided by its overall length in meters and multiplied by something called its permeability. It’s this term, represented by the Greek letter 𝜇, that is affected by the presence of this soft iron bar.

It so happens that the permeability of iron is greater than the permeability of air, which was in the core of the coil beforehand. But the important point is not how the permeability changes but rather that it changes. Because there’s a change in 𝜇, there’s a change in 𝐿. And because there’s a change in 𝐿, there’s a change in 𝑋 sub 𝑙. And because there’s a change in 𝑋 sub 𝑙, the inductive reactance, we’re no longer at resonance in our circuit. And because we’re not at resonance, the impedance of our circuit, 𝑍, is no longer equal simply to 𝑅.

We now have to include all three terms. We can write then that when there is an iron bar in the inductor coil, the maximum current in our circuit is equal to 𝑉 sub max divided by the square root of 𝑅 squared plus the quantity 𝑋 sub 𝑙 minus 𝑋 sub 𝑐 squared. And recall that what we want to do is make a comparison between the maximum current values with and without this iron bar.

In both sides, our numerator, 𝑉 sub max, is the same value, so the comparison will come down to the denominator, the impedance. Since the resistance, 𝑅, is less than the square root of 𝑅 squared plus the quantity 𝑋 sub 𝑙 minus 𝑋 sub 𝑐 squared, that means that the overall fraction, 𝐼 sub max, when there is no iron bar is greater than 𝐼 sub max when there is an iron bar. This all tells us that the insertion of the soft iron bar makes current in a circuit decrease.

If we go back to our answer choices, we see that that’s an option. It’s option b. With the insertion of the soft iron bar, the reading of the ammeter will decrease.

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