Overview

Transcript

0:00Introduction

Hello everyone! One small puzzle first. In a series circuit on an A C supply, voltmeters read eighty volts on the resistor, one hundred on the inductor and forty on the capacitor. So the source gives two hundred and twenty volts, right? Hmm, hold that thought; Example two will settle it. This chapter has three sets of JEE Main style problems; let's see how to use them.

0:27How this chapter works

So, EMI, electromagnetic induction, and A C, alternating current, usually give two or three questions in a JEE Main shift. Okay? The usual types: an e m f from a changing flux or a moving rod, then an L R or L C circuit, L for inductor, R resistor, C capacitor, right? Or an LCR, resonance or transformer question. Now tell me, our voltmeter puzzle, which type is it? Yes, exactly, a series LCR circuit, which is Set three, alternating current! Set one is induction and motional e m f, and Set two is inductance, L R and L C circuits. Clear? Real JEE Main questions, from twenty twenty one to twenty twenty six, are at the end of Chapters one to five. Here, each set has fifteen original problems. With me?

1:25Set 1 tools: Faraday, Lenz and motional emf

Now, Set one. Capital phi, the Greek letter, is the flux: field B times area capital A times cos theta, with theta from the normal, not from the plane. Okay? Faraday's law: the curly E, the e m f, is minus N times d phi over d t, for N turns. The charge q is N times the size of delta phi, over R, the resistance; delta, the little triangle, means the change, and the bars mean its size. It is independent of the time taken. Right? In the figure, a rod slides on rails joined by a resistor R; the crosses show the field B, into the page, and B's arrow marks a vector. For length l and velocity v, the e m f is B v l, when B, v and l are mutually perpendicular, which the upside-down T sign shows; otherwise use the perpendicular components. Got it? The orange arrows show the current, I, going up through the rod. So tell me, why up, and not down? It is Lenz's law, yes! The rod moves right, the loop grows, so the flux into the page rises. The induced current opposes this rise with a field out of the page, so it flows anticlockwise, up the rod. Under the figure, this current is B l v over R, and the force F, B squared l squared v over R, opposes the motion. With me?

2:49Set 2 tools: inductance, LR and LC circuits

Now, Set two. N phi equals L times I, with L the self-inductance. The back e m f is minus L times d I over d t, and the stored energy U is one half L I squared. For two coils with mutual inductance M, the e m f in coil two is minus M times d I one over d t; the small one and two label the coils. Okay? In an L R circuit, the current grows as curly E over R, times one minus e to the power minus t over tau; small e is the base of natural logarithms. Tau, the Greek letter, is L over R, the time constant, and the decay is i nought times e to the power minus t over tau, with i nought the starting current. Right? Now let me ask you this: at the instant of switch-on, with no current yet, what does the inductor behave like? Like an open circuit, very good! An inductor's current cannot jump, so for that instant it stays zero. In the steady state, an ideal inductor is just a plain wire. Clear? In L C oscillations, omega, the angular frequency, is one over the root of L times C, and i max, the maximum current, is omega times q nought, the maximum charge. Got it?

4:07Set 3 tools: AC circuits and transformers

Now, Set three. For sinusoidal A C only, I rms, the root mean square value, is the peak, I nought, over root two, and meters read rms values. Okay? X L, the inductor's reactance, is omega L, and the voltage across it leads the current by ninety degrees, the small circle meaning degrees. X C, the capacitor's reactance, is one over omega C, and the current leads the voltage across it by ninety degrees. Right? In series, the impedance, capital Z, is the root of R squared plus X L minus X C, the whole squared. At resonance X L equals X C, so omega nought is one over root L C, and Z equals R, its least value. Clear? Average power P is V rms times I rms times cos phi, where cos phi, R over Z, is the power factor; phi is the phase difference. With me? And an ideal transformer: V s over V p equals N s over N p, the turns on the secondary and primary. It works by mutual induction, so only with alternating current. Got it?

5:20Example 1: emf when flux is 4t squared minus 3t mWb

Example one, an easy one from Set one. The flux through a loop is phi equals four times t squared, minus three times t, in milliweber, m W b. Find the e m f at t equals one second. Pause the video and try it first. Done? Okay, let's check. At one second, the flux is four minus three, which gives one milliweber. So the e m f is one millivolt, written m V, right? No, no, wait, that is the trap! The e m f depends on how fast the flux changes, not on the flux itself. So differentiate: d phi over d t is eight times t, minus three, in milliweber per second. Okay? At one second, eight minus three gives five, and one milliweber per second is one millivolt. So the e m f is five millivolts in size. Right? Now the direction. d phi over d t is greater than zero, so the flux is rising. By Lenz's law, the induced current's field opposes this rise, against the original field; that is the minus sign in Faraday's law. Clear?

6:32Example 2: series LCR source voltage and power factor

Example two, a medium one from Set three: our opening puzzle. In a series LCR circuit, V R, the V with a small R below, is eighty volts across the resistor; V L is one hundred and V C is forty volts. Find the source voltage and the power factor. Pause the video and try it first. Done? Okay, let's check. Eighty plus one hundred plus forty gives two hundred and twenty volts. Is that right? Oh no, careful, that is the trap! In the figure, V R, in orange, lies along the current I, V L up and V C down: they are phasors, not plain numbers. So, one hundred minus forty gives sixty volts, the dashed level marked V L minus V C, okay? That is at right angles to V R, so by Pythagoras, V is the root of eighty squared plus sixty squared, which gives one hundred volts. Not two hundred and twenty, and our puzzle is solved. Right? Now the power factor. Phi is the angle between the source voltage V and the current, the same phi as in the impedance triangle on the right. So cos phi is V R over V, eighty over one hundred, which gives zero point eight. Clear?

7:50Quick revision: spot the set, Lenz direction, phasor sum

So, quick revision. First, spot the set, then use that set's tools. Second, the e m f is the rate of change of flux, and Lenz's law gives its direction. Third, in A C, voltages in series add as phasors, not as plain numbers. Now open Set one and try its practice problems. All the best!