Charging and Discharging

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. We charge an empty capacitor from a cell through a resistor, and some energy becomes heat in the resistor. So a tiny resistor should mean tiny heat, right? Well, no, and that is the surprise. Exactly half the energy the cell supplies becomes heat, whatever the resistance. And does the capacitor ever get completely full? Questions on charging and discharging come up often in JEE, so let's begin.

0:31Charging: the charge climbs towards CE

So, look at the graph on the left: the charge q against time t. We charge a capacitor C through a resistor R, from a cell of e m f capital E, okay? Tau, the Greek letter on screen, equals R times C: the time constant. Right? Now tell me. The capacitor starts empty. Why does its charge level off at C E, the dashed line at the top? Yes, exactly, because the current dies away to zero! With no current there is no voltage across R, so the full e m f sits across the capacitor, and q heads towards C times E, the final charge. On the way, q equals C E, times one minus e to the power minus t over tau in brackets, where e is the base of natural logs. Clear? So after one time constant, when t equals tau, is the capacitor full? No, it is not full yet, and that surprises many students. Put t equals tau, and q is one minus e to the power minus one, times C E, about zero point six three C E; the wavy equals sign means approximately. And the curve only creeps towards the dashed line, so it is never completely full. Got it?

1:47Charging: the current dies away

Now the current. At switch-on, the capacitor has no charge, so no voltage across it. So what current flows at that instant? Yes, E over R, the largest it will ever be! Then it dies away: I equals E over R, times e to the power minus t over tau. Okay? Now the voltages. V C, that is V with a small C below, is the capacitor voltage, E times one minus e to the power minus t over tau. V R, the resistor voltage, is E times e to the power minus t over tau. Together they always add up to E, the e m f. With me? And a graph trick. Take the natural log, written l n on screen: the natural log of I equals the natural log of E over R, minus t over R C, so the log of I against time is a straight line, and its slope, minus one over R C, gives the time constant. Right?

2:47Discharging: the charge falls away

Now the right-hand graph. We remove the cell, and the capacitor discharges through the resistor, starting with charge q nought, q with a small zero below. Then q equals q nought times e to the power minus t over tau, and the current is q nought over tau times the same factor. Clear? After one time constant, q is e to the power minus one times q nought, about zero point three seven q nought. Thirty seven percent is left, as the graph marks. Got it? Now a tricky one. When the charge halves, does the stored energy halve at that same moment? No, and many students fall for this. The stored energy goes as q squared, so it falls as e to the power minus two times t over tau. When the charge halves, the energy is already down to a quarter. Okay? So the half-life of the charge is tau times the natural log of two, but the energy halves sooner, at one half of tau times the natural log of two. With me?

3:52Energy: half stored, half lost as heat

Now, back to our puzzle. Charging from zero, the cell pushes a charge C E through an e m f of E volts. So it does work C E times E, that is, C E squared. Right? And how much of that energy is stored in the capacitor at the end? Yes, one half C E squared, the energy of a capacitor charged to E volts. That is only half of it. Clear? So where did the other half go? Into heat in the resistor: C E squared minus one half C E squared, which leaves one half C E squared. And look, there is no R in the answer, so the heat is the same for any resistor. That solves our puzzle! Got it? More generally, take a capacitor from V one to V two with a cell of e m f V two; the small one and two label the voltages. The heat is one half C, times the square of V two minus V one. Okay?

4:50Example 1: time constant, times for 50% and 90% charge

Alright, example one. R is one megaohm, written M and the Greek letter capital omega, the ohm sign, and C is two microfarad, written mu F with the Greek letter mu. Find the time constant, and the times to reach fifty and ninety percent of the final charge. Pause the video and try it first. Done? Okay, let's check. Tau is R times C: ten to the power six ohm, times two times ten to the power minus six farad. That gives two seconds. Easy, right? Now, for a fraction f of the final charge, set one minus e to the power minus t over tau equal to f. The double arrow gives t equals tau times the natural log of one over one minus f. Clear? For fifty percent, f is a half, so one minus f is a half too. So t is two times the natural log of two, which gives about one point three nine seconds. Got it? Now ninety percent. Here f is zero point nine, so t is two times the natural log of one over zero point nine, right? No, and this is the trap! We need one over one minus f, which is one over zero point one, that is ten. So t is two times the natural log of ten, which gives about four point six seconds, longer than for fifty percent. With me?

6:18Example 2: time constant using R eq, final capacitor voltage

Now a harder one. An ideal ten volt cell is in series with R one, two kilo ohm, then a five microfarad capacitor in parallel with R two, also two kilo ohm; the small one and two label the resistors. Find the time constant and the final capacitor voltage. Pause the video and try it first. Done? Okay, let's check. Both resistors are in the circuit, so tau is C times R one plus R two, right? No, and that is exactly the trap. First find R e q, R with small letters e q below: the resistance the capacitor sees with every ideal cell replaced by a wire. Then R one and R two are in parallel, shown by the two vertical bars. Two times two, over two plus two, gives one kilo ohm. Okay? So tau is C times R e q: five times ten to the power minus six farad, times ten to the power three ohm. That gives five milliseconds, written m s on screen. Right? Now the final voltage. Long after switching on, what current flows through the capacitor? Yes, exactly, zero! In the steady state, a capacitor in a D C circuit carries no current, like a break in the wire. So current flows only through R one and R two. Clear? C is in parallel with R two, so V C equals V R two, the voltage across R two. The equal resistors share the ten volts equally, so that is ten times one half, which gives five volts. Got it?

7:58Quick revision: tau = RC, 63% and 37%, half energy as heat

So, quick revision. Charging, q equals C E times one minus e to the power minus t over tau; discharging, q nought times e to the power minus t over tau; and tau is R times C, okay? In one tau, sixty three percent charged or thirty seven percent left, and the energy halves at one half of tau times the natural log of two. From zero, half of C E squared is stored and half becomes heat, whatever the resistance. Now go and try the practice problems on this page. All the best!