Capacitors in DC Circuits

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Take a ten volt battery, two resistors and one capacitor, all in a single circuit. Close the switch, and the battery current is five ampere. Wait for a while, and the same battery now gives only two ampere. Same battery, same resistors, nothing touched. So what changed? Keep that question in mind, because by the end of this video you will see exactly what happened. Capacitors inside D C circuits come up often in JEE, and two simple pictures handle most of these questions. So, let's begin.

0:38Steady state: the capacitor branch is open

So, the first picture is the steady state: the circuit after a long time, when nothing changes any more. Now tell me, in the steady state, does any current flow into the capacitor, the capital C on screen? If you said none at all, you are right! In the steady state, no current flows into a capacitor. Clear? So the capacitor's branch acts as an open circuit, just like a wire that has been cut. Nothing flows anywhere along that branch. Right? Now the method, step by step. First, find the currents in the rest of the circuit, using the series and parallel rules you already know, okay? Then find the potential difference V across the capacitor's terminals. And the charge on it is q equals C times V, where C is the capacitance. Got it?

1:29The trap: a resistor in series with the capacitor

Now, here is a trap that catches many students. Suppose a resistor R sits in series with the capacitor, in the same branch. In the steady state, is there any voltage drop across that resistor? Let us see. The branch is open, so the current I through that resistor is zero. Zero current, even though the resistor is right there in the circuit. With me? And the drop across a resistor is I times R, written V R, with a small R below the V, okay? Zero times R is zero. So there is no potential drop across it at all, however large R is. Right? So the capacitor gets the full voltage across its branch, as if that resistor were not there. Many students subtract a drop across this resistor and get the wrong charge. So please, do not be one of them. Clear?

2:20Just after switching: the charge cannot jump

Now the second picture: the instant just after you close or open a switch. So tell me, can the charge on a capacitor change in zero time? It cannot, and that is the key idea. The charge q, and so the voltage, V equals q over C, cannot change instantly. Whatever it was just before the switch, it is the same just after. Okay? Now take an uncharged capacitor. Here t equals zero plus, with the small plus sign up top, means the instant just after the switch is closed. The voltage was zero before, so it is still zero at t equals zero plus. Current can flow, but with zero voltage across it, and that is exactly what a plain wire does. So an uncharged capacitor acts as a wire. Right? And a charged capacitor? At that instant it keeps its voltage, which is q over C, and does not drop to zero. So it acts like a cell, with an e m f, the electromotive force, equal to q over C, okay? Two pictures, and that is the whole concept. Got it?

3:26Topper's trick: redraw the circuit twice

Now a trick that toppers use. Don't try to handle a capacitor circuit in your head. Redraw it, twice. For the instant just after switching, draw each uncharged capacitor as a plain wire, and each charged one as a cell whose e m f is q over C, okay? For the steady state, draw the circuit again, and simply replace every capacitor by a gap, since no current flows into it. Then read the voltage V across each gap, which is the voltage of the capacitor that sat there. With me? Each time, you get an ordinary circuit of resistors and cells, which you already know how to solve. As simple as that!

4:05Example 1: steady charge and energy of a 3 uF capacitor

Now, Example one, an easy one. Here, a twelve volt ideal battery drives a two ohm and a four ohm resistor in series. The Greek letter omega on screen stands for ohm. Then a three micro farad capacitor is connected across the four ohm resistor; the Greek letter mu means micro, one millionth. Find the steady charge on the capacitor, and its energy. Pause the video and try it first. Done? Okay, let's check. It is the steady state, so the capacitor draws no current; its branch is open. So the same current flows through the two ohm and the four ohm resistors, a simple series circuit. Okay? The capacitor is across the four ohm resistor, so it has the same voltage, V four; the small four below the V marks the four ohm resistor. The total resistance is six ohm, and the twelve volts are shared in the ratio of the resistances, so V four is twelve times four over six, which gives eight volts. Right? Now the charge. q equals C times V, so three micro farad times eight volts gives twenty four micro coulomb. Clear? And the energy. From electrostatics, the energy stored, U, equals one half times C times V squared. Eight squared is sixty four. Three times sixty four is one hundred and ninety two, and half of that is ninety six. So the energy is ninety six micro joule. That's it!

5:36Example 2: battery current at t = 0 and in steady state

Now Example two, a little harder. And this is our puzzle from the start! Here, a ten volt ideal battery and a two ohm resistor are in series with a three ohm resistor, and an uncharged capacitor is connected in parallel with that three ohm resistor. Find the battery current just after the switch is closed, and in the steady state. Pause the video and try it first. Done? Okay, let's check. Start with the instant just after the switch is closed. The capacitor is uncharged. So what does it act as, at t equals zero plus? Yes, exactly, it acts as a plain wire! And that wire sits in parallel with the three ohm resistor, so it shorts it out: the current takes the wire, and none goes through the three ohm resistor. Okay? So only the two ohm resistor is left in the circuit. The battery current, I, is ten over two, which gives five ampere. Right? Now wait for a long time. In the steady state, no current flows into the capacitor, so its branch is open, and the three ohm resistor is back in action. With me? Now the two ohm and the three ohm resistors are simply in series, five ohm in total. Ten over five gives two ampere. See, puzzle solved! The current starts at five ampere and settles at two ampere, because the capacitor goes from a wire to an open gap. Clear?

7:04Quick revision: open in steady state, wire or cell at t = 0

So, quick revision. In the steady state, the capacitor branch is open; find the voltage across it, and then q equals C times V, okay? No current means no drop across a resistor in series with the capacitor. And just after switching, an uncharged capacitor is a wire, and a charged one is a cell whose e m f is q over C, remember? Now go and try the practice problem on this page. You will find it easy. All the best!