### 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.