Mixed Problems: Set 3 (Power and RC Circuits)

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a bulb rated one hundred watt at two hundred volts, and run it on one hundred and sixty volts. That is eighty percent of the voltage, so eighty percent of the power, eighty watt, right? Hold on to that answer. Today is our third set of mixed problems, on power and R C circuits. Problems like these come up often in JEE Main. So, let's begin.

0:28Idea 1: a rating fixes the resistance

Idea number one is our puzzle. Here, a one hundred watt, two hundred volt bulb is used on one hundred and sixty volts; W stands for watt and V for volt. Find its power. First, the power in a resistor. P equals V times I, which equals I squared R, which equals V squared over R, where V is the voltage, I the current, and R the resistance; the small two above means squared. Okay? The label on the bulb is its rating, a power at a voltage. Which of these three forms links the power and the voltage, with nothing else? Yes, exactly, V squared over R! Call the rating P nought at V nought; the small zero below marks the rated value. Then R equals V nought squared over P nought, and we take this resistance as constant. Right? So at any other voltage V, the power is P nought times V squared over V nought squared. Here that is one hundred times, one hundred and sixty over two hundred, the whole squared. With me? One hundred and sixty over two hundred is zero point eight, and zero point eight squared is zero point six four. One hundred times that gives sixty four watt, and not eighty watt! The power goes as the voltage squared, and that solves our puzzle. Clear?

1:45Idea 2: the most power from a source

Now, idea number two. Here, a ten volt source of internal resistance five ohm feeds a variable load; the capital omega on screen, the Greek letter, means ohm. Find the maximum power. In general, the source has e m f capital E and internal resistance small r, inside it, and the load is capital R, outside it. The load gets P equals E squared times capital R, over capital R plus small r, the whole squared. Okay? So tell me, which load takes the most power from this source? Yes, exactly, the load equal to the internal resistance, capital R equals small r! Then P max, the P with a small max below, equals E squared over four times small r. But the same current heats small r inside, so half the power is lost there; the efficiency, with the percent sign on screen, is only fifty percent. Got it? Now the numbers. E squared is ten squared, which is one hundred, and four times small r is four times five, which is twenty. One hundred over twenty gives five watt, when the load is five ohm. Right?

2:52Idea 3: a capacitor in the steady state

Now, idea number three, capacitors. Here, a nine volt ideal battery drives one ohm and two ohm in series, and a four microfarad capacitor is across the two ohm; mu F, with the Greek letter mu, means microfarad. Find its steady charge. Long after switching on, what current flows through the capacitor? Yes, exactly, zero! In the steady state, no current flows through C, the capacitor, so its branch acts like an open circuit. Okay? So the current flows only through the two resistors in series. Nine over, one plus two, gives three ampere, the letter A on screen. Right? The capacitor sits across the two ohm resistor, so it has the same voltage. V C, the V with a small C below, is the voltage across the two ohm, three times two, which gives six volts. Clear? And the charge q equals C times V C, that is, four microfarad times six volts. That gives twenty four microcoulomb, written mu C on screen. With me?

3:56Example 1: time for capacitor to reach 75% of final charge

Alright, example one. R is two megaohm, written capital M omega, and C is five microfarad. How long does the capacitor take to reach seventy five percent of its final charge? Pause the video and try it first. Done? Okay, let's check. First, the time constant. Tau, the Greek letter on screen, is R times C, the resistance times the capacitance. Two times ten to the power six ohm, times five times ten to the power minus six farad, gives ten seconds, the small s on screen. Okay? Now, seventy five percent of the charge, so three quarters of the time constant, seven point five seconds, right? No, no, careful, the charge does not grow in a straight line! While charging, q equals C E, times one minus e to the power minus t over tau, where capital E is the e m f of the cell and e is the base of natural logs. We want q to be a fraction f of the final charge C E, with f equal to zero point seven five. Clear? Solve for t with the natural log, written l n. Then t equals tau times the natural log of one over, one minus f. One minus zero point seven five is zero point two five, and one over that is four. So t equals ten times the natural log of four. With me? The natural log of four is about one point three eight six, so t is about thirteen point nine seconds; the wavy equals sign means approximately. That is about one point four time constants, much more than three quarters of one. Got it?

5:34Example 2: energy of discharging capacitor after one tau

Now a harder one. Here, a ten microfarad capacitor at fifty volts discharges through a resistor. Find its energy at the start, and after one time constant. Pause the video and try it first. Done? Okay, let's check. The energy at the start, U nought, the U with a small zero below, is one half C V squared, where C is the capacitance and V the voltage. That is one half, times ten times ten to the power minus six, times fifty squared. Okay? Fifty squared is two thousand five hundred. One half, times ten, times two thousand five hundred, gives twelve thousand five hundred, times ten to the power minus six, so U nought is twelve point five times ten to the power minus three joule. That is twelve point five millijoule, written m J on screen. With me? Now, after one time constant, thirty seven percent of the charge is left. So thirty seven percent of the energy is left too, right? No, no, wait, that is the trap! The charge falls as q nought, the starting charge, times e to the power minus t over tau. But the stored energy goes as q squared; the sign like a stretched alpha means proportional to. So the energy falls as e to the power minus two times t over tau, twice as fast. Clear? Put t equals tau, and the energy is twelve point five times e to the power minus two. That power of e is about zero point one three five, so the energy is about one point six nine millijoule. So only about thirteen and a half percent of the energy is left, much less than thirty seven percent. Got it?

7:10Quick revision: P goes as V squared, max power, RC circuits

So, quick revision. The rating fixes the resistance, R equals V nought squared over P nought, so the power goes as V squared. Maximum power, E squared over four times small r, comes at capital R equals small r, and in the steady state no current flows through a capacitor. To reach a fraction f of the final charge takes tau times the natural log of one over, one minus f, and while discharging, the energy falls as e to the power minus two times t over tau. You did really well today. Now go and try the practice problem on this page. All the best!