Capacitors in Steady-State Circuits

Transcript

0:00Introduction

Hello everyone! Let me start with a small puzzle. Three capacitors are joined at one common point, and their other ends are held at ten, twenty and thirty volts. So, what is the potential of that common point? Twenty volts, the simple average? Hold that thought, we will settle it by the end of this video. Capacitor circuits come up often in JEE, and four simple rules handle most of them. So, let's begin.

0:28Rule 1: the steady state

So, rule number one. Picture a circuit with a battery, some resistors and a capacitor. We switch it on and wait a long time. Now tell me, how much current flows through the branch that has the capacitor? None at all, the current is zero. Once the capacitor has charged up, its charge stops changing, and no current flows through its branch. We call this the steady state. Okay? And here is a trap. Suppose a resistor sits in the same branch, in series with that capacitor. The current I through it is zero, that is the I equals zero on screen, so current times resistance is zero too. There is no voltage drop across that resistor at all, and many students forget this. Right?

1:14Rule 2: isolated plates keep their charge

Now, rule number two. Some plates in a circuit are joined only to each other, by a wire, and to nothing else: no battery, and no earth. We call such a set of plates isolated. Okay? Now let me ask you this. Before we switch on, these plates carry no charge. After we switch on, what is their total charge? Still zero, exactly! Charge cannot cross the gap of a capacitor, and no wire brings any in from outside. So the total charge on an isolated set of plates stays constant, and usually that constant is zero. Clear? That is why two capacitors in series, with an isolated junction between them, carry equal charges: plus q on one middle plate and minus q on the other, where q stands for charge. With me? But careful here. If a battery is connected to that junction, it is not isolated any more, and the two charges need not be equal. So always check whether the junction is isolated. Got it?

2:15Rule 3: the loop rule

Rule number three is the loop rule. The voltage across a capacitor is q over C, its charge over its capacitance. Okay? Now go around a closed loop made of capacitors and batteries only. The voltage drops across the capacitors, taken with proper signs, add up to the e m f, the electromotive force, of the batteries in that loop. If a resistor in the loop carries current, its I R drop must be added too. On screen, the big Greek sigma means sum, and the curly Greek epsilon is the e m f of each battery. So, the sum of q over C equals the sum of epsilon. Right?

2:54Rule 4: the node method

And now rule number four, the node method, the key to our puzzle. Several capacitors meet at one point, called a node, and this node is isolated. Their far ends sit at known potentials. C i is the capacitance of each capacitor, and V i is the potential at its far end; the small i below is a subscript, just a counting label. Okay? So, what is the total charge on the plates joined there? Zero, yes, because the node is isolated, and that is rule two! Each plate at the node carries C i times V node minus V i, where the word node, written small below, is again a subscript. So the sum of C i, times V node minus V i, equals zero. With me? Now solve this for V node. V node equals the sum of C i times V i, over the sum of C i. Write this down, it is worth it. Clear? Look at its shape. It is a weighted average of the potentials, each one weighted by its capacitance. So a bigger capacitor pulls the node closer to its own potential. Got it?

4:04Example 1: steady charge on a 3 microfarad capacitor

Now, Example one. Here, a three microfarad capacitor is in parallel with a six ohm resistor. This pair is in series with a three ohm resistor, across a nine volt battery. On screen, the Greek mu means micro, and the capital Greek omega is the ohm. Find the charge on the capacitor in steady state. Pause the video and try it first. Done? Okay, let's check. It is the steady state, so rule one says no current flows through the capacitor. So the whole current goes through the six ohm resistor and then the three ohm resistor, in series. Okay? So the current I equals nine over six plus three. That is nine over nine, which gives one ampere, the capital A on screen. Right? Now, the battery gives nine volts, so the capacitor charges up to the full nine volts. Then the charge is three times nine, twenty seven microcoulomb, right? Oh no, careful, that is the trap! The capacitor is in parallel with the six ohm resistor, so it has the same voltage as that resistor. That is one ampere times six ohm, which gives six volts. Clear? So the charge Q equals C times V: three microfarad times six volts gives eighteen microcoulomb. Not twenty seven, but eighteen. Easy, once you remember rule one. Got it?

5:29Example 2: node potential and charges, ends at 10, 20, 30 V

And now Example two, our puzzle from the start. Capacitors of one, two and three microfarad are joined at a common point O, and their other ends are at ten, twenty and thirty volts. Find V O, the potential of point O, where the small O is a subscript, and find the charge on each capacitor. Pause the video and try it first. Done? Okay, let's check. Many students take the simple average of ten, twenty and thirty, and say twenty volts. Is that right? No, and that is exactly the trap! Point O is isolated, so we use the node method, the weighted average. On top, one times ten, plus two times twenty, plus three times thirty, that is ten plus forty plus ninety. Below, one plus two plus three, which is six. So one hundred and forty over six gives seventy over three volts, about twenty three point three volts. More than twenty, because the three microfarad capacitor at thirty volts pulls the hardest. Right? Now the charges. Each capacitor has V i minus V O across it, so the plate at its far end, the outer plate, carries C i times V i minus V O, okay? For the one microfarad capacitor, we get one times ten minus seventy over three. Ten is thirty over three, so this gives minus forty over three microcoulomb. With me? For the two microfarad capacitor, we get two times twenty minus seventy over three. Twenty minus seventy over three is minus ten over three, and two times that gives minus twenty over three microcoulomb. Right? And for the three microfarad capacitor, three times thirty minus seventy over three. Thirty minus seventy over three is twenty over three, and three times that gives exactly twenty microcoulomb. Clear? Now a quick check. The plates at point O carry the opposite charges. So what must those charges add up to? Zero, because point O is isolated! Forty over three plus twenty over three is sixty over three, which is twenty, and twenty minus twenty gives zero. So rule two checks our answer. Beautiful, isn't it?

7:51Quick revision: steady state, isolated plates, loop, node

So, quick revision. In steady state, no current flows through a capacitor branch, so a resistor in that branch has no voltage drop. Isolated plates keep their total charge, and around a loop of capacitors and batteries, the sum of q over C equals the sum of the e m f. And at an isolated node, V node is the weighted average: the sum of C i times V i, over the sum of C i. Now go and try the practice problems on this page. All the best!