Mixed Problems: Set 2 (Inductance, LR and LC Circuits)

Transcript

0:00Introduction

Hello everyone! One small puzzle first. Take a twelve volt battery, a four ohm resistor in series, and an ideal coil with a twelve ohm resistor across it. Close the switch, and the battery current is zero point seven five amperes. Then, with nothing touched, it settles at three amperes. So who changed the current? Hmm, the coil did something, right? This is Set two of our mixed problems, on inductance, L R and L C circuits, and these come up often in JEE Main. Let's begin.

0:36Tool 1: self and mutual inductance

So, our tool number one is inductance. When the current, capital I, in a coil changes, the coil makes a back e m f: curly E equals minus L times d I over d t. The minus sign is Lenz's law: it opposes the change. L, the self-inductance, is in henry, the capital H in brackets. Okay? Now let me ask you this. If the current in a coil doubles, does its self-inductance double too? No, it does not, careful! L depends only on the coil's geometry and the medium inside it. For a long solenoid, L equals mu nought times N squared times capital A, over small l. Mu nought, with a small zero below, is the permeability of free space, N the number of turns, capital A the area, and small l the length. Right? The coil also stores energy in its magnetic field: U equals one half times L times I squared. See the square on the current. Clear? When the current I one in coil one changes, a second coil gets an e m f: curly E two equals minus M times d I one over d t, with small ones and twos written below. M, the mutual inductance, equals k times the square root of L one times L two, where k, the coupling coefficient, lies from zero to one, as the less than or equal signs show. Got it? Join these coupled coils in series, and careful, the total is not just L one plus L two! L equals L one plus L two, plus or minus two M: plus when the fluxes aid each other, and minus when they oppose. With me?

2:20Tool 2: LR circuits, growth and decay

Tool number two, the L R circuit: a coil L and a resistor R, with a battery of e m f curly E, okay? Tau, the Greek letter, is the time constant, L over R, and i nought, small i with a zero below, equal to curly E over R, is the final, steady current. Right? Connect the battery, and the current grows: i equals i nought times one minus e to the power minus t over tau, with e the base of natural logarithms. After one time constant, the rising curve reaches about zero point six three times i nought. Clear? Remove the battery and close the circuit on itself. The current decays from its starting value i nought: i equals i nought times e to the power minus t over tau, the falling curve, down to about zero point three seven times i nought after one time constant. Okay? Now suppose the coil was carrying no current. At the very instant we close the switch, what is the current through it? Zero, yes, exactly! The current through an inductor cannot change suddenly, so just after a switch is thrown, the inductor keeps its earlier current. Got it? If that current was zero, the double arrow says it acts as an open circuit. And in the steady state the current stops changing, so an ideal inductor is just a plain wire. Example two needs both. With me?

3:47Tool 3: LC oscillations

Now tool number three, L C oscillations. Connect a charged capacitor C across an ideal coil L, with no resistance. The charge q swings between q nought and minus q nought, the solid curve, with angular frequency omega, one over the square root of L times C, okay? The total energy stays fixed: the capacitor's q squared over two C, plus the coil's one half L times i squared, equals q nought squared over two C, and also one half L times i max squared. Here i max, with max written small below, is the peak of the dashed curve. Right? So i max is omega times q nought. And q nought is C times V nought, where V nought is the capacitor's starting voltage. So i max is omega times C times V nought, which equals V nought times the square root of C over L, nice and short. Clear? Now here is a quick trap for you. The charge oscillates with some frequency, call it f. At what frequency does the energy in the capacitor oscillate? Most students say f, but careful, it is two f! The energy goes as q squared, so it peaks at plus q nought and again at minus q nought, twice a cycle. Got it?

5:06Example 1: maximum current, 4 uF capacitor on a 10 mH coil

Now our first example. So, a four microfarad capacitor, charged to fifty volts, is connected to a ten millihenry coil. Find the maximum current. On screen, mu F is microfarad, m H is millihenry, and capital V is volt. Pause the video and try it first. Done? Okay, let's check. First, S I units: micro is ten to the power minus six, so C is four times ten to the power minus six farads, with capital F for farad. Milli is ten to the power minus three, so L is ten times ten to the power minus three, which is ten to the power minus two henry. Right? Now, which of our tools gives the maximum current straight from the starting voltage? Energy conservation, yes, exactly! i max equals V nought times the square root of C over L, okay? C over L is four times ten to the power minus six, over ten to the power minus two, which gives four times ten to the power minus four. Clear? Its square root is zero point zero two, so i max is fifty times zero point zero two, which gives one ampere, with capital A for ampere. Forget the micro and the milli, and you are far off. Okay?

6:28Example 2: battery current just after switch-on and steady

Now Example two, our puzzle from the start. So, a twelve volt battery is connected through four ohms to an ideal inductor in parallel with twelve ohms; the omega on screen means ohm. Find the battery current just after switching on, and in the steady state. Pause the video and try it first. Done? Okay, let's check. Many students say the inductor is a wire, so it shorts the twelve ohm resistor, and the current is twelve over four, three amperes, straight away. Is that correct? No, that is exactly the trap, so be careful! Just after switching on, the inductor current cannot jump from zero, so the inductor is an open circuit. Got it? So the current flows through four ohms and twelve ohms in series. Twelve over the sum of four and twelve, that is twelve over sixteen, gives zero point seven five amperes. Clear? Now the steady state. The current has stopped changing, so the ideal inductor acts as a plain wire, and it shorts the twelve ohm resistor. Right? Only the four ohm resistor is left, and twelve over four gives three amperes. So the puzzle is solved: zero point seven five amperes at first, settling at three amperes. Three amperes was right, but only in the steady state! With me?

7:53Quick revision: L dI/dt, time constant L/R, LC peak current

So, quick revision. Inductance: the back e m f is minus L times d I over d t, the energy is one half L times I squared, and coupled coils in series add plus or minus two M, okay? L R circuits: tau is L over R, an open circuit just after a switch if no current flowed, and a plain wire when steady. L C circuits: omega is one over root L C, and i max is V nought times root C over L, clear? Now try the practice problems on this page. All the best!