Kirchhoff's Laws

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a twelve volt cell and a six volt cell, each with a two ohm resistor, positive terminals at the top, connected side by side across a four ohm resistor. Each cell pushes current out of its positive terminal, right? Hmm, not so fast, because one of these two cells is actually being charged! So which one is it? Two simple rules from Kirchhoff will settle it. These circuits come up often in JEE, so let's begin.

0:32Rule 1: the junction rule

So, rule number one, the junction rule. Now, a junction is a point where several wires meet, and the currents entering it add up to the currents leaving it. On screen, capital I stands for current, the Greek capital sigma means sum, and the small words in and out below mean entering and leaving. Clear? But why must this be true? Suppose more current entered a junction than left it. What would happen there? Charge would keep piling up there, yes, exactly! But charge is conserved, and in the steady state no charge piles up at a junction. So the junction rule is simply conservation of charge. Got it?

1:11Rule 2: the loop rule

Now rule number two, the loop rule. Walk round any closed loop and add every change in potential, rises as plus and drops as minus. This algebraic sum is zero. The Greek capital delta, the little triangle, means change, and capital V stands for potential, so sigma delta V is the sum of the potential changes. With me? Now tell me, why must this sum be zero? Remember, you come back to the point you started from. Same point, same potential, exactly! So all the rises and drops cancel. In energy terms, a charge taken round the loop has the same energy when it comes back, so the net energy it gains from the cells equals the energy it loses in the resistors. So the loop rule is conservation of energy. Okay?

2:02Signs as you walk along a branch

Now the signs, where mistakes often happen. In a resistor, current flows from higher potential to lower potential. So if you walk through a resistor along the current, does the potential rise or drop? Yes, it drops, because you are going downhill in potential. So for a resistor R carrying a current I, the change is minus I times R, right? And walking against the current, you go uphill, so the change is plus I times R, clear? Now a cell, with an e m f, capital E, okay? The minus and plus signs on screen are its negative and positive terminals. Going from negative to positive, the potential rises, so plus E, right? And from positive to negative, it falls, so minus E, okay? Now careful, this is a common trap. For a cell, this sign depends only on which terminal you enter first, not on the direction of the current. Got it?

2:59Potential difference along a branch

Now use these signs along a branch, from point A to point B, okay? Start with V A, the potential at point A, written V with a small A below. Subtract I times R for each resistor crossed along the current, add it when you cross against the current, and for each cell add E with the sign from our rule, which is what the plus minus sign means. You arrive at V B, the potential at point B, right? So it is just a walk. Start with V A, add every rise and every drop, and you arrive at V B, the end point. Clear?

3:33The node-potential method

For bigger circuits, here is a neat method, the node potential method, where a node is a junction point. Step one: choose any node, and call its potential zero. Right? Step two: call the unknown node potentials V one, V two, and so on, which is what the small numbers and the three dots mean. Okay? Step three: at each unknown node, write the junction rule as, the sum of the currents leaving the node equals zero. Clear? Now here is the current in each branch. Take a branch from node X to node Y, shown by the arrow, with resistance R and an e m f E pointing towards Y, so its positive terminal faces Y, okay? The current from X to Y equals V X minus V Y plus E, all over the resistance R, right? Step four: solve these linear equations, and every current then follows. Got it?

4:29Example 1: VA - VB across a 4 ohm resistor and a 10 V cell

Alright, example one. Here, a current of one ampere flows from point A to point B, through a four ohm resistor and then through a ten volt cell, from its positive terminal to its negative terminal. The capital A after the one means ampere, the capital V after the ten means volt, and the Greek letter omega means ohm. Find V A minus V B, the potential difference. Pause the video and try it first. Done? Okay, let's check. Start at point A and walk along the current. First the resistor, so minus I times R, minus one times four, where the cross means times. That gives minus four volts. Right? Now the cell. Many students say a cell always gives energy, so plus ten volts. What do you say, plus ten or minus ten? It is minus ten, careful here! We go through the cell from its positive terminal to its negative terminal, so the potential falls by ten volts, whatever the current does. Clear? So the walk reads, V A, minus four, minus ten, equals V B, okay? Rearrange, and V A minus V B equals fourteen volts. One wrong sign would spoil it all, so always walk carefully.

5:47Example 2: 12 V and 6 V cells in parallel, which is charged

Now a harder one, and this is our puzzle from the start. Nodes X, at the top, and Y, at the bottom, are joined by three branches. Branch one has a twelve volt cell, positive terminal towards X, and two ohms. Branch two is a four ohm resistor. Branch three has a six volt cell, positive terminal also towards X, and two ohms. Find all the currents. Pause the video and try it first. Done? Okay, let's check with the node method. Take the bottom node as zero, so V Y is zero, and call the top node's potential V, just one unknown, okay? Now the junction rule at X, the top node. By our branch formula, branch one carries twelve minus V, over two, up into the node. Branch three carries six minus V, over two, also upwards. Branch two carries V over four, down and out. Current in equals current out. With me? Multiply by four, and the left side becomes thirty six minus four times V, which equals V itself. So thirty six equals five times V, and V is seven point two volts. Okay? Now the currents. Branch one: twelve minus seven point two, over two, gives two point four amperes, upwards. Branch two: seven point two over four, gives one point eight amperes, downwards. Clear? Now branch three: six minus seven point two, over two, gives minus zero point six amperes, a negative current. So tell me, did we make a mistake somewhere? No, there is no mistake at all! The minus sign only says the current flows the other way, so zero point six amperes flows down, into the positive terminal of the six volt cell. So the six volt cell is being charged, and our puzzle is solved! Check the junction too: two point four in, one point eight plus zero point six out, so it all adds up.

7:47Quick revision: junction rule, loop rule, sign convention

So, quick revision. Junction rule: current in equals current out, because charge is conserved. Loop rule: potential changes round a closed loop add up to zero, because energy is conserved. Signs: minus I times R along the current, plus E from negative to positive, and a negative current simply flows the other way. Now go and try the practice problems on this page. You can do it. All the best!