Part C: Concept Checks

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Put a resistor, a coil and a capacitor in series on an A C supply, with a voltmeter across the coil alone. Can it read more than the supply voltage itself? Surely not, right? Keep your answer in mind. Today we do eight concept checks on induction and alternating current. They come up often in JEE, so let's begin.

0:27Check 1: does a changing flux always give a current?

So, check one. Faraday's law: curly E, the emf, equals minus N times d phi over d t, with N the number of turns and phi, the Greek letter, the magnetic flux. The minus sign is Lenz's law: the induced emf opposes the change in flux. The double arrow means which gives: a changing flux gives an emf. Okay? Now here is the check. Does a changing flux always produce a current? Not always, and this is where students slip. The emf is always there, but a current flows only in a closed circuit; then capital I, the current, equals curly E over R, with R the resistance. Right? Now the second question. Suppose a loop moves, without rotating, inside a uniform magnetic field. Is a current induced? No current, that is correct. The arrow on B means a vector. In a uniform field, the loop keeps the same field and the same angle, so the flux stays constant: no net emf round the loop, and no current. Clear?

1:32Check 2: is the induced electric field conservative?

Now check two. In the figure, the grey crosses are a field B into the page, inside the shaded circle, and it is increasing; we skip the graph. It induces an electric field: the closed integral of E dot d l equals minus d phi over d t. The circle on the integral means once round a loop, the arrows mean vectors, and d l is a small step along it. With me? Here is the check. Is this induced electric field conservative, like the field of a charge at rest? No, it is not, and the figure shows why. The orange circles, the E lines, are closed loops, going anticlockwise as Lenz's law says, to oppose the increase in the flux into the page. Okay? So the work done in carrying a charge once round such a loop is not zero; the crossed equals sign means not equal to. For a conservative field it is zero, so this field is not conservative.

2:29Check 3: power in a pure inductor or capacitor

Got it? Now check three, on A C power. The average power, capital P, is V rms times I rms times cos phi, with r m s meaning root mean square. Here phi is not the flux but the phase difference between the voltage and the current; cos phi is the power factor. Right? Here is the check. What is the average power consumed by a pure inductor, or a pure capacitor, on A C? Look at the phase first. In a pure inductor, L, the current lags the voltage by ninety degrees, the small circle meaning degrees. In a pure capacitor, C, it leads by ninety degrees. Clear? Either way, phi is ninety degrees, and cos ninety degrees is zero. So the average power is zero, even though a current flows. Okay?

3:18Check 4: what can a transformer not do?

Now check four, the transformer. For an ideal transformer, which works only on A C, V s over V p equals N s over N p: V for the voltage and N for the turns, small s for the secondary coil and small p for the primary. So it changes the voltage, and the current too. With me? So tell me two things. Does a transformer change the frequency of the supply? And can it increase the power? The answer is no to both. The secondary emf comes from the same changing flux in the core, so the frequency stays the same. And the output power cannot exceed the input; this sign means less than or equal to. Got it? Now a related check. Why is electric power sent over long lines at a high voltage? For the same power, a higher voltage, V, means a smaller current, capital I, okay? And the heat lost in the line is I squared R, with R the line's resistance. Because the current is squared, a smaller current gives a much smaller loss. Clear?

4:26Example 1: inductor on DC, just after closing and long after

Now, Example one, the easier one. How does an ideal inductor behave just after a switch is closed, and after a long time, in a D C circuit? Pause the video and try it first. Done? Okay, let's check. An ideal inductor has no resistance, so it is just a plain wire from the very start, right? Oh no, careful, that is the trap. The current through an inductor cannot change suddenly: it was zero before closing, so it is still zero just after. In the left graph, small i, the current, starts from zero at time small t equal to zero; we skip the rest. Okay? The battery is connected, yet no current flows. So at first, the inductor acts as an open circuit. With me? And after a long time? The dark blue curve flattens out: the current stops changing, so d i over d t is zero. V L, the V with a small L below, is the voltage across the coil, L, the self-inductance, times d i over d t, so it is zero: the inductor is a plain wire. Right? So the current settles at curly E, the battery's emf, over R, the resistance: the top dashed line, just as if the coil were not there. Very good!

5:43Example 2: can coil voltage in series LCR exceed the supply?

Now Example two, a little harder, and it is our puzzle from the start. Can the voltage across the inductor in a series LCR circuit be larger than the source voltage? Pause the video and try it first. Done? Okay, let's check. In a series circuit, the voltages across the parts add up to the supply voltage. So no part can be more than the whole, correct? Oh no, careful, on A C they add as phasors, not as plain numbers. In the figure, the current I is horizontal; V R, the V with a small R below, is the orange arrow along it, V L the blue arrow up, and V C the green arrow down. So V, the source voltage, is the square root of, V R squared plus, V L minus V C, the whole squared. Okay? Now look: V L and V C point opposite ways, so they are in antiphase. They largely cancel, and only the difference, the dashed level marked V L minus V C, enters the source voltage; we skip the triangle. Clear? Now take resonance, where X L, the inductive reactance, equals X C, the capacitive reactance. Then the impedance, capital Z, equals R, its least value, so the current, V over R, is the greatest. Got it? So V L and V C are each Q times the source voltage, where Q, the quality factor, is omega nought times L, over R, and omega nought, the omega with a small zero, is the resonant angular frequency. Right? So whenever Q is more than one, the greater-than sign, V L is larger than the source voltage: the answer is yes! That solves our puzzle: the voltmeter across the coil really can read more than the supply. Perfect!

7:23Quick revision: flux and current, AC power, transformer, LCR

So, quick revision. Changing flux always gives an emf, but a current only in a closed circuit, and the induced electric field is not conservative. Pure inductors and capacitors take zero average power; a transformer keeps the frequency and adds no power. An inductor on D C is open, then a wire, and in series LCR, V L can exceed the source voltage. Now try the other concept checks on this page. All the best!