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# Electric Circuits and Interconnection Laws

Module by: Don Johnson. E-mail the author

Summary: A circuit is the electrical/physical analogue of a system. Kirchoff's laws provde connection rules about "nodes" in a circuit to determine relations between various circuit elements.

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A circuit connects circuit elements together in a specific configuration designed to transform the source signal (originating from a voltage or current source) into another signal—the output—that corresponds to the current or voltage defined for a particular circuit element. A simple resistive circuit is shown in Figure 1. This circuit is the electrical embodiment of a system having its input provided by a source system producing v in t v in t .

To understand what this circuit accomplishes, we want to determine the voltage across the resistor labeled by its value R 2 R 2 . Recasting this problem mathematically, we need to solve some set of equations so that we relate the output voltage v out v out to the source voltage. It would be simple—a little too simple at this point—if we could instantly write down the one equation that relates these two voltages. Until we have more knowledge about how circuits work, we must write a set of equations that allow us to find all the voltages and currents that can be defined for every circuit element. Because we have a three-element circuit, we have a total of six voltages and currents that must be either specified or determined. You can define the directions for current flow and positive voltage drop any way you like. When two people solve a circuit their own ways, the values of their variables may not agree, but current and voltage values for each element will agree. Do recall in defining your voltage and current variables that the v-i relations for the elements presume that positive current flow is in the same direction as positive voltage drop. Once you define voltages and currents, we need six nonredundant equations to solve for the six unknown voltages and currents. By specifying the source, we have one; this amounts to providing the source's v-i relation. The v-i relations for the resistors give us two more. We are only halfway there.

What we need to solve every circuit problem are mathematical statements that express what the interconnection of elements is. Said another way, we need the laws that govern the electrical connection of circuit elements. First of all, the places where circuit elements attach to each other are called nodes. Two nodes are explicitly indicated in Figure 1; a third is at the bottom where the voltage source and resistor R 2 R 2 are connected. Electrical engineers tend to draw circuit diagrams—schematics— in a rectilinear fashion. Thus the long line connecting the bottom of the voltage source with the bottom of the resistor is intended to make the diagram look pretty. This line simply means that the two elements are connected together. Kirchoff's Laws, one for voltage and one for current, determine what a connection among circuit elements means. These laws can help us analyze this circuit.

## Kirchoff's Current Law

At every node, the sum of all currents entering a node must equal zero. What this law means physically is that charge cannot accumulate in a node; what goes in must come out. In the example, Figure 1, below we have a three-node circuit and thus have three KCL equations.

(i) i 1 =0 i i 1 0
(1)
i 1 i 2 =0 i 1 i 2 0
(2)
i+ i 2 =0 i i 2 0
(3)
Note that the current entering a node is the negative of the current leaving the node.

Given any two of these KCL equations, we can find the other by adding or subtracting them. Thus, one of them is redundant and, in mathematical terms, we can discard any one of them. The convention is to discard the equation for the (unlabeled) node at the bottom of the circuit.

### Exercise 1

In writing KCL equations, you will find that in an nn-node circuit, exactly one of them is always redundant. Can you sketch a proof of why this might be true? Hint: It has to do with the fact that charge won't accumulate in one place on its own.

#### Solution

KCL says that the sum of currents entering or leaving a node must be zero. If we consider two nodes together as a "supernode", KCL applies as well to currents entering the combination. Since no currents enter an entire circuit, the sum of currents must be zero. If we had a two-node circuit, the KCL equation of one must be the negative of the other, We can combine all but one node in a circuit into a supernode; KCL for the supernode must be the negative of the remaining node's KCL equation. Consequently, specifying n1 n 1 KCL equations always specifies the remaining one.

## Kirchoff's Voltage Law (KVL)

The voltage law says that the sum of voltages around every closed loop in the circuit must equal zero. A closed loop has the obvious definition: Starting at a node, trace a path through the circuit that returns you to the origin node. KVL expresses the fact that electric fields are conservative: The total work performed in moving a test charge around a closed path is zero. The KVL equation for our circuit is

v 1 + v 2 v=0 v 1 v 2 v 0
(4)
In writing KVL equations, we follow the convention that an element's voltage enters with a plus sign if traversing the closed path, we go from the positive to the negative of the voltage's definition.

For the example circuit above, we have three v-i relations, two KCL equations, and one KVL equation for solving for the circuit's six voltages and currents.

v= v in v v in
(5)
v-i:
v 1 = R 1 i 1 v 1 R 1 i 1
(6)
v out = R 2 i out v out R 2 i out
(7)
KCL:
(i) i 1 =0 i i 1 0
(8)
i 1 i out =0 i 1 i out 0
(9)
KVL:
v+ v 1 + v out =0 v v 1 v out 0
(10)

We have exactly the right number of equations! Eventually, we will discover shortcuts for solving circuit problems; for now, we want to eliminate all the variables but v out v out . The KVL equation can be rewritten as v in = v 1 + v out v in v 1 v out . Substituting into it the resistor's v-i relations, we have v in = R 1 i 1 + R 2 i out v in R 1 i 1 R 2 i out . Yes, we temporarily eliminate the quantity we seek. Though not obvious, it is the simplest way to solve the equations. One of the KCL equations says i 1 = i out i 1 i out , which means that v in = R 1 i out + R 2 i out =( R 1 + R 2 ) i out v in R 1 i out R 2 i out R 1 R 2 i out . Solving for the current in the output resistor, we have i out = v in R 1 + R 2 i out v in R 1 R 2 . We have now solved the circuit: We have expressed one voltage or current in terms of sources and circuit-element values. To find any other circuit quantities, we can back substitute this answer into our original equations or ones we developed along the way. Using the v-i relation for the output resistor, we obtain the quantity we seek.

V out = R 2 R 1 + R 2 v in V out R 2 R 1 R 2 v in
(11)

## Exercise 2

Referring back to Figure 1, a circuit should serve some useful purpose. What kind of system does our circuit realize and, in terms of element values, what are the system's parameter(s)?

### Solution

The circuit serves as an amplifier having a gain of R 2 R 1 + R 2 R 2 R 1 R 2 .

The results shown in other modules (circuit elements,KVL and KCL, interconnection laws) with regard to the circuit above (Figure 1), and the values of other currents and voltages in this circuit as well, have profound implications.

Resistors connected in such a way that current from one must flow only into another—currents in all resistors connected this way have the same magnitude—are said to be connected in series. For the two series-connected resistors in the example, the voltage across one resistor equals the ratio of that resistor's value and the sum of resistances times the voltage across the series combination. This concept is so pervasive it has a name: voltage divider.

The input-output relationship for this system, found in this particular case by voltage divider, takes the form of a ratio of the output voltage to the input voltage.

v out v in = R 2 R 1 + R 2 v out v in R 2 R 1 R 2
(12)
In this way, we express how the components used to build the system affect the input-output relationship. Because this analysis was made with ideal circuit elements, we might expect this relation to break down if the input amplitude is too high (Will the circuit survive if the input changes from 1 volt to one million volts?) or if the source's frequency becomes too high. In any case, this important way of expressing input-output relationships—as a ratio of output to input—pervades circuit and system theory.

The current i 1 i 1 is the current flowing out of the voltage source. Because it equals i 2 i 2 , we have that v in i 1 = R 1 + R 2 v in i 1 R 1 R 2 :

## Resistors in series:

The series combination of two resistors acts, as far as the voltage source is concerned, as a single resistor having a value equal to the sum of the two resistances.
This result is the first of several equivalent circuit ideas: In many cases, a complicated circuit when viewed from its terminals (the two places to which you might attach a source) appears to be a single circuit element (at best) or a simple combination of elements at worst. Thus, the equivalent circuit for a series combination of resistors is a single resistor having a resistance equal to the sum of its component resistances.

Thus, the circuit the voltage source "feels" (through the current drawn from it) is a single resistor having resistance R 1 + R 2 R 1 R 2 . Note that in making this equivalent circuit, the output voltage can no longer be defined: The output resistor labeled R 2 R 2 no longer appears. Thus, this equivalence is made strictly from the voltage source's viewpoint.

One interesting simple circuit (Figure 6) has two resistors connected side-by-side, what we will term a parallel connection, rather than in series.

Here, applying KVL reveals that all the voltages are identical: v 1 =v v 1 v and v 2 =v v 2 v . This result typifies parallel connections. To write the KCL equation, note that the top node consists of the entire upper interconnection section. The KCL equation is i in i 1 i 2 =0 i in i 1 i 2 0 . Using the v-i relations, we find that

i out = R 1 R 1 + R 2 i in i out R 1 R 1 R 2 i in
(13)

## Exercise 3

Suppose that you replaced the current source in Figure 6 by a voltage source. How would i out i out be related to the source voltage? Based on this result, what purpose does this revised circuit have?

### Solution

Replacing the current source by a voltage source does not change the fact that the voltages are identical. Consequently, v in = R 2 i out v in R 2 i out or i out = v in R 2 i out v in R 2 . This result does not depend on the resistor R 1 R 1 , which means that we simply have a resistor ( R 2 R 2 )across a voltage source. The two-resistor circuit has no apparent use.

This circuit highlights some important properties of parallel circuits. You can easily show that the parallel combination of R 1 R 1 and R 2 R 2 has the v-i relation of a resistor having resistance 1 R 1 +1 R 2 1= R 1 R 2 R 1 + R 2 1 R 1 1 R 2 1 R 1 R 2 R 1 R 2 . A shorthand notation for this quantity is R 1 R 2 R 1 R 2 . As the reciprocal of resistance is conductance, we can say that for a parallel combination of resistors, the equivalent conductance is the sum of the conductances.

Similar to voltage divider for series resistances, we have current divider for parallel resistances. The current through a resistor in parallel with another is the ratio of the conductance of the first to the sum of the conductances. Thus, for the depicted circuit, i 2 = G 2 G 1 + G 2 i i 2 G 2 G 1 G 2 i . Expressed in terms of resistances, current divider takes the form of the resistance of the other resistor divided by the sum of resistances: i 2 = R 1 R 1 + R 2 i i 2 R 1 R 1 R 2 i .

Suppose we want to pass the output signal into a voltage measurement device, such as an oscilloscope or a voltmeter. In system-theory terms, we want to pass our circuit's output to a sink. For most applications, we can represent these measurement devices as a resistor, with the current passing through it driving the measurement device through some type of display.

In circuits, a sink is called a load; thus, we describe a system-theoretic sink as a load resistance R L R L . Thus, we have a complete system built from a cascade of three systems: a source, a signal processing system (simple as it is), and a sink.

We must analyze afresh how this revised circuit, shown in Figure 9, works. Rather than defining eight variables and solving for the current in the load resistor, let's take a hint from other analysis (series rules, parallel rules). Resistors R 2 R 2 and R L R L are in a parallel configuration: The voltages across each resistor are the same while the currents are not. Because the voltages are the same, we can find the current through each from their v-i relations: i 2 = v out R 2 i 2 v out R 2 and i L = v out R L i L v out R L . Considering the node where all three resistors join, KCL says that the sum of the three currents must equal zero. Said another way, the current entering the node through R 1 R 1 must equal the sum of the other two currents leaving the node. Therefore, i 1 = i 2 + i L i 1 i 2 i L , which means that i 1 = v out (1 R 2 +1 R 1 ) i 1 v out 1 R 2 1 R 1 .

Let R eq R eq denote the equivalent resistance of the parallel combination of R 2 R 2 and R L R L . Using R 1 R 1 's v-i relation, the voltage across it is v 1 = R 1 v out R eq v 1 R 1 v out R eq . The KVL equation written around the leftmost loop has v in = v 1 + v out v in v 1 v out ; substituting for v 1 v 1 , we find

v in = v out ( R 1 R eq +1) v in v out R 1 R eq 1
(14)
or
v out v in = R eq R 1 + R eq v out v in R eq R 1 R eq
(15)

Thus, we have the input-output relationship for our entire system having the form of voltage divider, but it does not equal the input-output relation of the circuit without the voltage measurement device. We can not measure voltages reliably unless the measurement device has little effect on what we are trying to measure. We should look more carefully to determine if any values for the load resistance would lessen its impact on the circuit. Comparing the input-output relations before and after, what we need is R eq R 2 R eq R 2 . As R eq =1 R 2 +1 R L 1 R eq 1 R 2 1 R L 1 , the approximation would apply if 1 R 2 1 R L 1 R 2 1 R L or R 2 R L R 2 R L . This is the condition we seek:

## Voltage measurement:

Voltage measurement devices must have large resistances compared with that of the resistor across which the voltage is to be measured.

## Exercise 4

Let's be more precise: How much larger would a load resistance need to be to affect the input-output relation by less than 10%? by less than 1%?

### Solution

R eq = R 2 1+ R 2 R L R eq R 2 1 R 2 R L . Thus, a 10% change means that the ratio R 2 R L R 2 R L must be less than 0.1. A 1% change means that R 2 R L <0.01 R 2 R L 0.01 .

## Example 1

We want to find the total resistance of the example circuit. To apply the series and parallel combination rules, it is best to first determine the circuit's structure: What is in series with what and what is in parallel with what at both small- and large-scale views. We have R 2 R 2 in parallel with R 3 R 3 ; this combination is in series with R 4 R 4 . This series combination is in parallel with R 1 R 1 . Note that in determining this structure, we started away from the terminals, and worked toward them. In most cases, this approach works well; try it first. The total resistance expression mimics the structure:

R T = R 1 ( R 2 R 3 + R 4 ) R T R 1 R 2 R 3 R 4
(16)
R T = R 1 R 2 R 3 + R 1 R 2 R 4 + R 1 R 3 R 4 R 1 R 2 + R 1 R 3 + R 2 R 3 + R 2 R 4 + R 3 R 4 R T R 1 R 2 R 3 R 1 R 2 R 4 R 1 R 3 R 4 R 1 R 2 R 1 R 3 R 2 R 3 R 2 R 4 R 3 R 4
(17)
Such complicated expressions typify circuit "simplifications". A simple check for accuracy is the units: Each component of the numerator should have the same units (here Ω3 Ω 3 ) as well as in the denominator ( Ω2 Ω 2 ). The entire expression is to have units of resistance; thus, the ratio of the numerator's and denominator's units should be ohms. Checking units does not guarantee accuracy, but can catch many errors.

Another valuable lesson emerges from this example concerning the difference between cascading systems and cascading circuits. In system theory, systems can be cascaded without changing the input-output relation of intermediate systems. In cascading circuits, this ideal is rarely true unless the circuits are so designed. Design is in the hands of the engineer; he or she must recognize what have come to be known as loading effects. In our simple circuit, you might think that making the resistance R L R L large enough would do the trick. Because the resistors R 1 R 1 and R 2 R 2 can have virtually any value, you can never make the resistance of your voltage measurement device big enough. Said another way, a circuit cannot be designed in isolation that will work in cascade with all other circuits. Electrical engineers deal with this situation through the notion of specifications : Under what conditions will the circuit perform as designed? Thus, you will find that oscilloscopes and voltmeters have their internal resistances clearly stated, enabling you to determine whether the voltage you measure closely equals what was present before they were attached to your circuit. Furthermore, since our resistor circuit functions as an attenuator, with the attenuation (a fancy word for gains less than one) depending only on the ratio of the two resistor values R 2 R 1 + R 2 =1+ R 1 R 2 1 R 2 R 1 R 2 1 R 1 R 2 1 , we can select any values for the two resistances we want to achieve the desired attenuation. The designer of this circuit must thus specify not only what the attenuation is, but also the resistance values employed so that integrators—people who put systems together from component systems—can combine systems together and have a chance of the combination working.

Figure 11 summarizes the series and parallel combination results. These results are easy to remember and very useful. Keep in mind that for series combinations, voltage and resistance are the key quantities, while for parallel combinations current and conductance are more important. In series combinations, the currents through each element are the same; in parallel ones, the voltages are the same.

## Exercise 5

Contrast a series combination of resistors with a parallel one. Which variable (voltage or current) is the same for each and which differs? What are the equivalent resistances? When resistors are placed in series, is the equivalent resistance bigger, in between, or smaller than the component resistances? What is this relationship for a parallel combination?

### Solution

In a series combination of resistors, the current is the same in each; in a parallel combination, the voltage is the same. For a series combination, the equivalent resistance is the sum of the resistances, which will be larger than any component resistor's value; for a parallel combination, the equivalent conductance is the sum of the component conductances, which is larger than any component conductance. The equivalent resistance is therefore smaller than any component resistance.

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