Combinations of Capacitors

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. You have two capacitors, of three microfarad and six microfarad, and you join them together. So what is the total capacitance? Three plus six, nine microfarad, right? Hmm, that is not always true! Join them one way, and you do get nine. Join them the other way, and you get just two microfarad, less than either of them. Combinations of capacitors come up often in JEE circuit questions, so let's find out why.

0:32Parallel: the same V across each

So, first the parallel connection. Here the capacitors sit side by side, all joined across the same two battery terminals. So every capacitor has the same potential difference, capital V, across it. Okay? Now, each capacitor stores its own charge. So let me ask you this: what total charge must the battery supply? Yes, exactly, the sum of the two! Capital Q equals Q one plus Q two, and each charge is C times V, so C one times V, plus C two times V, where C is the capacitance. The small one and two below are just labels. Right? Take V common, and the equivalent capacitance, capital C, is C one plus C two, and so on; the three dots mean and so on. In parallel, capacitances simply add. Clear? And since each charge is C times the same V, charges divide in proportion to capacitance: Q one is to Q two as C one is to C two, and the colon means is to. The bigger capacitor takes the bigger charge. Got it?

1:40Series: the same Q on each

Now the series connection, where capacitors are joined one after another in a chain. Charge cannot jump the gap inside a capacitor, so the plates joined in the middle can only share charge with each other. So every capacitor carries the same charge, capital Q, okay? So now it is the voltages that add. V equals V one plus V two, which is Q over C one, plus Q over C two. With me? Divide both sides by capital Q, and one over C equals one over C one, plus one over C two, and so on. In series the reciprocals add, and one warning: after adding, do not forget to flip. Clear? Now a topper's trick for exactly two capacitors. C equals C one times C two, over C one plus C two, the product over the sum. Right? The battery voltage is shared between the two capacitors. So tell me, which one gets the larger share, the bigger capacitor or the smaller one? Most students say the bigger one. No, no, careful, it is the opposite! Each V is Q over C, with the same Q, so the smaller capacitor takes the larger voltage. V one equals C two over C one plus C two, times V; see, C two sits on top, so voltages divide inversely.

3:00Parallel versus series: at least four times

Okay? Now compare the two connections for the same pair. C parallel over C series equals C one plus C two, the whole squared, over C one times C two; parallel and series below the C are labels. And this ratio is greater than or equal to four, the sign with the line under it. Clear? So let me ask you this: for which pair of capacitors is the ratio exactly four? C one plus C two, the whole squared, minus four times C one times C two, equals C one minus C two, the whole squared, which is zero only when C one equals C two. So two equal capacitors give exactly four, and any other pair gives more. Got it?

3:44The balanced Wheatstone bridge

Now a famous arrangement, the Wheatstone bridge. One arm has C one followed by C two, the other has C three followed by C four, and a bridge capacitor joins the middle points of the two arms. With me? The bridge capacitor gets charge only if there is a potential difference across it. So when would it get no charge? Yes, exactly, when both middle points are at the same potential! Each arm splits the voltage inversely as its capacitors, so if C one over C two equals C three over C four, both arms split it the same way. That is a balanced bridge. Okay? Since Q equals C times V, zero potential difference means zero charge. So the bridge capacitor does nothing, and you can simply remove it. Clear?

4:32Many plates, joined alternately

One last arrangement. Take n equally spaced plates, and join them alternately to the two terminals: the odd plates to one side, the even plates to the other. Okay? Each neighbouring pair is a capacitor of epsilon naught times A, over d. Epsilon naught, the epsilon with a small zero, is the permittivity of free space, while A is the plate area, and d is the spacing. Count the gaps, not the plates: n plates give n minus one capacitors, written in brackets on screen. Right? Each of these has one plate on each terminal, so all have the same V across them, which means they are in parallel. So capital C equals n minus one, times epsilon naught times A, over d. Got it?

5:18Example 1: charge and voltage, 3 and 6 microfarad in series

Now our first example, which brings back our puzzle. Three microfarad and six microfarad, in series, are connected to ninety volts; mu F, with the Greek letter mu, is microfarad, and the V after ninety is volts. Find the charge, and the voltage across each. Pause the video and try it first. Done? Okay, let's check. Product over sum, where the cross sign means times: three times six over three plus six is eighteen over nine, which gives two microfarad. That solves our puzzle: in series, three and six give just two. Clear? Now the charge: Q equals C times V, two times ninety, which is one hundred and eighty microcoulombs; mu C is microcoulombs. And in series, each capacitor carries this same charge. Right? So which capacitor gets the larger voltage, the three microfarad or the six microfarad? Yes, the smaller one, just as we saw! Across the three microfarad, V one is one hundred and eighty over three, which gives sixty volts, and across the six microfarad, V two is one hundred and eighty over six, which gives thirty volts. And sixty plus thirty gives ninety, the battery voltage. Got it?

6:39Example 2: equivalent capacitance of a balanced bridge

Now a harder one. Capacitors of one, two, two and four microfarad form a Wheatstone bridge, one and two in one arm, two and four in the other, with five microfarad across the middle. Find the equivalent capacitance between the ends. Pause the video and try it first. Done? Okay, let's check. First, the five microfarad in the middle: is it in series with something, or in parallel? It is neither, and that is the trap! One over two and two over four are both one half, so the bridge is balanced, and the five microfarad carries no charge. We remove it. Okay? Now each arm is just two capacitors in series. For arm one, one times two, over one plus two, gives two thirds of a microfarad. Right? For arm two, two times four is eight, and two plus four is six, so eight over six gives four thirds of a microfarad. Clear? Both arms run between the same two ends, so they are in parallel, and we add. Two thirds plus four thirds gives two microfarad, and the five microfarad never mattered.

7:54Quick revision: series, parallel, bridge and n plates

Got it? So, quick revision. Parallel: the same V, capacitances add, and charge goes in proportion to capacitance. Series: the same Q, reciprocals add, and the smaller capacitor takes the larger voltage. Balanced bridge: remove the middle capacitor; and n plates give n minus one, times epsilon naught times A, over d. Now go and try the practice problems on this page. All the best!