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.

To start on our solution, let’s draw a diagram of the circuit. In this exercise, we have a circuit with an inductor, a capacitor, and a resistor arranged in series with an AC power supply. We’re told further that the circuit is at resonance.

For an RLC circuit, like we have here, the resonant frequency, 𝑓 sub zero, is equal to one over two 𝜋 times the square root of the inductor value times the capacitor value. Given this circuit in this condition, we want to know what effect inserting an iron bar into the inductor coil will have on the current that runs throughout this circuit.

Let’s first consider the circuit current before the iron bar is inserted into the coil of the inductor. For a circuit with these components, we can write a modified version of Ohm’s law, where instead of 𝑉 equals 𝐼𝑅 we now say 𝑉 equals 𝐼 times 𝑍, where 𝑍 is the circuit impedance. Impedance is a concept that includes resistance, like we might expect, and it also includes the reactance of the inductor and the capacitor in the circuit.

The inductive reactance, 𝑋 sub 𝐿, and the capacitive reactance, 𝑋 sub 𝐶, are given in equation form, with 𝑋 sub 𝐿 being two 𝜋 times the frequency times the inductance, 𝐿, and 𝑋 sub 𝐶, the capacitive reactance, being equal to one over two 𝜋 times the frequency of the circuit multiplied by its capacitance.

When we consider 𝑋 sub 𝐿, 𝑋 sub 𝐶, and 𝑍, we’ll see something interesting happens when we apply the fact that our circuit is in resonance. Starting with 𝑋 sub 𝐿, we see it’s equal to two 𝜋 times the resonant frequency, one over two 𝜋, times the square root of 𝐿𝐶, all multiplied by 𝐿. The factors of two 𝜋 cancel out. And we’re left with a simplified expression of the square root of 𝐿 over 𝐶, where 𝐿 is the inductance of our inductor without an iron core, 𝐶 is the capacitance of our capacitor, and, as we’ll use in a bit, 𝑅 is the resistance of our resistor.

Moving on to considering 𝑋 sub 𝐶, the capacitive reactance, this is equal to one over two 𝜋 times the resonant frequency, one over two 𝜋, times the square root of 𝐿 times 𝐶, all multiplied by the capacitance, 𝐶. Once again, we see the factors of two 𝜋 cancel out. And the denominator of this fraction simplifies to the square root of 𝐶 over 𝐿, so that, overall, 𝑋 sub 𝐶 is also equal to the square root of 𝐿 over 𝐶.

Here’s why this is important that 𝑋 sub 𝐶 and 𝑋 sub 𝐿 are equal. When we go to calculate the impedance, 𝑍, of our circuit, its total resistance to the flow of current, we see that it’s equal to the square root of 𝑅 squared plus 𝑋 sub 𝐿 minus 𝑋 sub 𝐶 quantity squared. But since 𝑋 sub 𝐿 is equal to 𝑋 sub 𝐶, this term is equal to zero. This means that, at resonance, the impedance of our circuit simplifies to the resistance, 𝑅, of the circuit, that is, the value of the circuit’s resistor.

Referring back to our modified expression for Ohm’s law, we can say then that the total current running through this circuit when it’s at resonance — we’ll call this 𝐼 sub 𝑅 — is equal simply to 𝑉 over 𝑅, where 𝑉 is the average voltage supplied by our AC source.

We’ve seen how the condition of resonance in our circuit has created a special value for the impedance, 𝑍. If the frequency of our circuit were at anything other than resonance, this term 𝑋 sub 𝐿 minus 𝑋 sub 𝑍 [𝐶] may well not be zero, and then the impedance overall would grow.

In fact, if we were to create a graph showing current versus frequency in the circuit, we will see that the current experiences a peak value at the resonant frequency 𝑓 sub zero. If our frequency would shift to either side of that resonant frequency, the overall circuit current would go down.

Returning to our inductor in our circuit diagram, we now imagine taking an iron core and filling up the coil of the inductor with this cylindrical core. The question is, what effect will this core have on the overall inductance of our inductor and therefore on the overall current in the circuit?

The inductance of our inductor increases thanks to the metal core in its coil, so that the final inductance of the inductor, 𝐿 sub 𝑓, will be greater than its initial inductance when it had just an air core. Looking at our equation for resonant frequency, we see that it depends on the value of the inductor, 𝐿. Since our circuit was at resonance when our inductor had a value simply of 𝐿 before the core was inserted, that means that when we increase the value of 𝐿 by inserting a core, we move the frequency of oscillation of our circuit off of the resonant frequency, 𝑓 sub zero.

According to our chart then, the current in the circuit will fall off to either side of the maximum experienced at resonance. We can show this mathematically too by observing that when our system is not in resonance, 𝑋 sub 𝐿 and 𝑋 sub 𝐶, the inductive and capacitive reactances, respectively, no longer cancel one another out.

This means that this expression for the final current in our circuit after the core has been inserted into the inductor gives us a value of current less than the value of 𝑉 over 𝑅, because 𝐼 sub 𝑓 has a larger denominator. We see then that the overall current in our circuit will decrease after we insert an iron rod into the inductor. This is answer option b.

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