Prism Glow - Electronics - EEES - Chapter 1 - D-C Circuits - 1-4 Fundamental Circuit Laws; Kirchhoff's Laws

Chapter 1-4 Fundamental Circuit Laws; Kirchhoff's Laws

Kirchhoff's Laws provide a systematic way to study, understand and analyze basic electric circuits. They follow naturally given the properties we have discussed in the previous three sections. There are two such laws. The first one is:

1. The algebraic sum of all the currents directed toward a junction are equal to zero.

The first law is a reference to the fact that, "TANSTAAFL". (If you do not know what this stands for, then you need to read Robert A. Heinlein's, excellent book, "The Moon is a Harsh Mistress".)

In other words, some currents may flow to a junction node, and some may flow from that same node, but in total no current will be left unaccounted for when you consider the directions of flows of each current and algebraically add them. Every ampere of current going in to a node will be matched by the same amount of current leaving the node.

As current flows through a circuit and encounters a node, the current will branch off down the various paths leaving the node. The amounts of current that flow down the different branches is based upon the relative resistances of the branches.

Let's look at a simple illustration. Here we have a circuit with 20 ohms

of resistance in total and a 30 volts

source. By Ohm's Law we can calculate that there ought to be 1.5

of current.

Illustration of current split and Kirchhoff's Law

And indeed there is as we see just past R1 where we inserted an ammeter. But what happens when we come to the node where R2 and R3 branch off from the main circuit? R2 is a 20 ohms

resistance and R3 is a 5 ohms

resistance. The total current splits into two parts. And the split is based upon the relative resistance of the two branches.

We can look at the next two ammeters, respectively in series with R2 and R3. We see that the higher resistance of R2 allows only .3 amperes

of current to flow. And the lower resistance in R3 allows 1.2 amperes

. The sum of .3 amperes

and 1.2

equals the total 1.5 amperes

sent in to the parallelled resistors.

Once the currents pass through the resistors in parallel, they recombine at terminal

and continue around the circuit. The fourth ammeter in the circuit registers 1.5 amperes

, showing that all current that entered the dual branches of the parallel resistors also left them, and sums to the original total.

Now we come to the second of Kirchhoff's laws:

2. The algebraic sum of all the voltage drops around a closed circuit is zero.

The second law is similar in that it is an accounting function. It says that given a circuit path, as you go around it, in either direction, the sum of the voltage drops across the various components and connectors will sum to zero when algebraically added to the voltage rise provided by a source.

Voltage drops across resistances are always in the same direction as the circuit current flow.

What does that mean? Recall Ohm's Law.

Voltage equals the product of the current and resistance. When a current passes through a resistance, energy transfer is effected. This energy transfer is known as Work. Evidence of this work having been done is the potential difference across the two sides of the resistance.

Another way of saying this is that there is a voltage drop across the resistance. Since the current has to pass through the resistance to cause work to occur, then (relative to a fixed reference point) the voltage will be higher on the side of the resistance where the current enters than on the side where it exits.

Voltage rises (or drops) across sources are always in the same direction, but are accounted for based upon the direction that the circuit path is followed.

Now that seems even more confusing that the rule for the voltage drops across resistances. But it is really quite simple. When you look at a schematic diagram for a DC circuit, each source will have a positive and a negative terminal. There is always a voltage rise when looking at the source from the perspective of going from the negative terminal to the positive terminal. But, were you to be analyzing the circuit and traversing it such that the path went from the positive to the negative terminal of the source, you would count this as a negative voltage gain, or a voltage drop. This is true regardless of which direction the current was flowing. Let's have a look at what this means with an illustration that expands upon the previous one.

Illustration of voltage drops and Kirchhoff's Law

Illustration of voltage drops and Kirchhoff's Law

We put a voltmeter across each resistance and across the source. We have the current going clockwise in this circuit. Therefore, there should be a voltage drop across each resistance. And there is. Note the polarity of the voltmeters across the resistances. What they are measuring is the voltage at the positive terminal (where the current enters) of each resistance in relation to the voltage at the negative terminal. The voltage is always higher where the current enters, so the meters are registering voltage drops.

The voltage across the source is a voltage rise from negative to positive. It is exactly opposite in sign but equal in magnitude to the combined voltage drops of the resistances in the circuit. Going around the circuit we have 30 volts

-15

-6

-9

= 0

, which is exactly what we should have to prove Kirchhoff's second law.

Electronics > EEES > Chapter 1 - DC Circuits > 1-4 Fundamental Circuit Laws; Kirchhoff's Laws
Chapter 1-4 Fundamental Circuit Laws; Kirchhoff's Laws Kirchhoff's Laws provide a systematic way to study, understand and analyze basic electric circuits. They follow naturally given the properties we have discussed in the previous three sections. There are two such laws. The first one is: 1. The algebraic sum of all the currents directed toward a junction are equal to zero. The first law is a reference to the fact that, "TANSTAAFL". (If you do not know what this stands for, then you need to read Robert A. Heinlein's, excellent book, "The Moon is a Harsh Mistress".) In other words, some currents may flow to a junction node, and some may flow from that same node, but in total no current will be left unaccounted for when you consider the directions of flows of each current and algebraically add them. Every ampere of current going in to a node will be matched by the same amount of current leaving the node. As current flows through a circuit and encounters a node, the current will branch off down the various paths leaving the node. The amounts of current that flow down the different branches is based upon the relative resistances of the branches. Let's look at a simple illustration. Here we have a circuit with 20 of resistance in total and a 30 source. By Ohm's Law we can calculate that there ought to be 1.5 of current. And indeed there is as we see just past R1 where we inserted an ammeter. But what happens when we come to the node where R2 and R3 branch off from the main circuit? R2 is a 20 resistance and R3 is a 5 resistance. The total current splits into two parts. And the split is based upon the relative resistance of the two branches. We can look at the next two ammeters, respectively in series with R2 and R3. We see that the higher resistance of R2 allows only .3 of current to flow. And the lower resistance in R3 allows 1.2 . The sum of .3 and 1.2 equals the total 1.5 sent in to the parallelled resistors. Once the currents pass through the resistors in parallel, they recombine at terminal and continue around the circuit. The fourth ammeter in the circuit registers 1.5 , showing that all current that entered the dual branches of the parallel resistors also left them, and sums to the original total. Now we come to the second of Kirchhoff's laws: 2. The algebraic sum of all the voltage drops around a closed circuit is zero. The second law is similar in that it is an accounting function. It says that given a circuit path, as you go around it, in either direction, the sum of the voltage drops across the various components and connectors will sum to zero when algebraically added to the voltage rise provided by a source. Voltage drops across resistances are always in the same direction as the circuit current flow. What does that mean? Recall Ohm's Law. Voltage equals the product of the current and resistance. When a current passes through a resistance, energy transfer is effected. This energy transfer is known as Work. Evidence of this work having been done is the potential difference across the two sides of the resistance. Another way of saying this is that there is a voltage drop across the resistance. Since the current has to pass through the resistance to cause work to occur, then (relative to a fixed reference point) the voltage will be higher on the side of the resistance where the current enters than on the side where it exits. Voltage rises (or drops) across sources are always in the same direction, but are accounted for based upon the direction that the circuit path is followed. Now that seems even more confusing that the rule for the voltage drops across resistances. But it is really quite simple. When you look at a schematic diagram for a DC circuit, each source will have a positive and a negative terminal. There is always a voltage rise when looking at the source from the perspective of going from the negative terminal to the positive terminal. But, were you to be analyzing the circuit and traversing it such that the path went from the positive to the negative terminal of the source, you would count this as a negative voltage gain, or a voltage drop. This is true regardless of which direction the current was flowing. Let's have a look at what this means with an illustration that expands upon the previous one. We put a voltmeter across each resistance and across the source. We have the current going clockwise in this circuit. Therefore, there should be a voltage drop across each resistance. And there is. Note the polarity of the voltmeters across the resistances. What they are measuring is the voltage at the positive terminal (where the current enters) of each resistance in relation to the voltage at the negative terminal. The voltage is always higher where the current enters, so the meters are registering voltage drops. The voltage across the source is a voltage rise from negative to positive. It is exactly opposite in sign but equal in magnitude to the combined voltage drops of the resistances in the circuit. Going around the circuit we have 30 -15 -6 -9 = 0 , which is exactly what we should have to prove Kirchhoff's second law.