Mixed Problems: Set 3 (Alternating Current)

Transcript

0:00Introduction

Hello everyone! Here is a real problem. Your bulb is rated sixty watts, one hundred and twenty volts, but the mains give two hundred and forty volts. Something in series must keep its current down to the rated value. Should it be a resistor, or a coil? And how big a coil? Keep your answer in mind; example two settles it. This is Set three of our mixed problems, on alternating current, which comes up often in JEE Main. So, let's begin.

0:30Idea 1: reactance, and who leads

Idea one is reactance, the opposition of a coil or a capacitor. X L, the X with a small L below it, is omega times L, and omega, the Greek letter, is two pi f, where f is the frequency. In the figure, orange is the current I, and blue the voltage, capital V, okay? In a resistor they are in phase; in the inductor, V is ninety degrees ahead, the small circle meaning degrees, so the current lags. Right? And X C is one over omega C, or one over two pi f C, okay? Here the current leads: the orange arrow is ahead. The word civil, under the figure, helps: in a capacitor, I leads V, and V leads I in an inductor. Clear? Now the question: twenty millihenry, the m H, at fifty hertz, the H z, and fifty microfarad, at one hundred hertz; mu, the Greek letter, means micro. Before using the formulas, what must you do to the twenty and the fifty? Convert them to S I units, yes, exactly! Twenty millihenry is zero point zero two henry, so X L is two pi, times fifty, times zero point zero two, which gives six point two eight ohms, the Greek capital omega for ohm. Okay? Fifty microfarad is five times ten to the power minus five farad. So X C is one over two pi, times one hundred, times that, which gives about thirty one point eight ohms. Forget the micro, and you are off by a million times! Got it?

2:10Idea 2: impedance and phase

Idea two is impedance. Here the resistance R is twelve ohms, X L is twenty and X C is four, in series on a one hundred and twenty volt A C supply, the capital V meaning volts. Find the current and its phase. Many students just add: twelve plus twenty plus four, thirty six ohms. Is that right? No, careful, that is the trap! In the figure, V L, in blue, points up, and V C, in green, points down, so they partly cancel. What is left is at right angles to V R, in orange, along the current I; together they make the source voltage, the black arrow. So in the impedance triangle on the right, capital Z, the impedance, is the square root of R squared, plus X L minus X C, the whole squared. And phi, the Greek letter, is the phase angle, with tan phi equal to X L minus X C, over R, okay? So X L minus X C is sixteen, and Z is the square root of twelve squared plus sixteen squared, which is the root of four hundred, so twenty ohms. Right? Like Ohm's law, I equals V over Z, one hundred and twenty over twenty, which gives six amperes, the capital A on screen. Clear? Tan phi is sixteen over twelve, or four over three. The double arrow means gives, and the wavy equals sign means approximately, so phi is about fifty three degrees. So, does the current lead or lag? It lags, yes, exactly! X L is bigger than X C, so the circuit is inductive, and the current lags by fifty three degrees. Got it?

3:57Idea 3: average power and the power factor

Idea three is power. The average power P is V rms times I rms times cos phi, where rms, in small letters, means root mean square. It also equals I rms squared times R, and cos phi, which is R over Z, is the power factor. Okay? Remember, A C meters read rms values, so one hundred volts A C means V rms is one hundred volts. And V rms times I rms alone is the apparent power, in volt-ampere, the V A on screen. Right? So, the question: five amperes from one hundred volts, at a power factor of zero point five. One hundred times five, so five hundred watts, right? Not quite; that is only the apparent power, in volt-ampere. Times the power factor, zero point five, it gives two hundred and fifty watts, the capital W on screen. Clear? Now the first half of our puzzle. In a pure inductor, phi is ninety degrees, so cos phi is zero, and no power is used. So a choke, a coil of small resistance, wastes little power, while a resistor wastes I squared R as heat. So the coil wins! Got it?

5:17Example 1: current, power factor and power of a coil on AC

Alright, example one. Take a coil of resistance eight ohms and reactance six ohms, connected to a one hundred volt A C supply. Find the current, the power factor and the power. Pause the video and try it first. Are you done? Okay, let's check. Resistance and reactance are at right angles, so Z is the square root of eight squared plus six squared, which gives ten ohms. So the current is one hundred over ten, which gives ten amperes. Right? The power factor, cos phi, is R over Z, eight over ten, which gives zero point eight. Clear? So the power is V rms times I rms times cos phi, one hundred times ten times zero point eight, which gives eight hundred watts. With me? Now check it the other way. Only the resistance uses power, so I squared R is ten squared times eight, which gives eight hundred watts again. Got it?

6:18Example 2: choke coil to run a 120 V bulb on 240 V mains

Now, our puzzle with numbers. Here a sixty watt, one hundred and twenty volt bulb must run on two hundred and forty volt, fifty hertz mains, with a series choke of negligible resistance. Find its inductance. Pause the video and try it first. Are you done? Okay, let's check. At one hundred and twenty volts the bulb takes sixty watts, so its current is sixty over one hundred and twenty, which gives zero point five ampere. Its resistance, one hundred and twenty over zero point five, is two hundred and forty ohms. Okay? On the mains, the same zero point five ampere must flow, so Z must be two hundred and forty over zero point five, which gives four hundred and eighty ohms. Right? Now the choke. Its reactance is four hundred and eighty minus two hundred and forty, so two hundred and forty ohms, right? No, no, wait, that is the same trap! Resistance and reactance are at right angles, so they add like the sides of the impedance triangle. So X L is the square root of four hundred and eighty squared, minus two hundred and forty squared, which gives two hundred and forty times root three, about four hundred and sixteen ohms. With me? Finally, since X L is two pi f times L, L is X L over two pi f. Two pi times fifty is about three hundred and fourteen, and four hundred and sixteen over three hundred and fourteen gives about one point three two henry, the capital H on screen. And that settles our puzzle. Clear?

8:01Quick revision: reactance, impedance and average power

So, quick revision. X L is omega L, and the current lags; X C is one over omega C, and the current leads; use S I units first. Z is the square root of R squared plus X L minus X C, the whole squared: add as phasors, not as plain numbers. And power is V rms times I rms times cos phi, or I rms squared times R; only the resistance uses it. Now go and try the practice problems on this page. All the best!