In deciphering the behavior of circuits from schematic diagrams, it is often useful to look at an isolated portion of a bigger circuit. In these instances, it is helpful to think about the current or voltage at one point of the circuit in order to reason what the current or voltage will be at some other point. It is a bit like finding one’s way on a road map. One focuses on the pathways between one’s present location and another point of interest and the obsta- cles in between. There is actually a rule known as Thevenin’s theorem that supports this sort of simplification [1]. Thevenin’s theorem states that for the purpose of analysis, any combination of voltage sources and resistors can be replaced by a single voltage source and one resistor in series. This theorem will be applied later in the discussion of the circuit in Figure 7.1.
There are a few other rules or laws that help to simplify this task even further [1]. The first of these ideas is that at any junction of wires in a circuit diagram, the total current going into the junction must equal the total current leaving the junction. This rule is sometimes referred to as Kirchhoff’s current law. Figure 7.1 shows a simple circuit with a battery and three resistors. Current flows from the positive end of the battery through the resistors to the negative end of the battery. Kirchhoff’s current law indicates that the current coming into point A from resistor,R1, equals the sum of the two currents leaving that point through resistorsR2 and R3. Kirchhoff also stated a rule about voltages. That idea is actually a declaration that voltage is a state function. That is, the rule states that the voltage or energy spent in going between any two points in a loop is the same regardless of which path is
1000 Ω
R2
2000 Ω R3
3000 Ω
Ground 0 V
A R1
B Vbat
+5 V + –
FIGURE 7.1 Simple circuit for demonstrating Kirchhoff’s and Ohm’s laws and Thevenin’s theorem.
These rules state: (i) The sum of currents going into any point, such as point A, is equal to the sum of the current leaving that point. (ii) The voltage between points A and B is the same regardless of the path taken between those points. (iii) The voltage difference between the leads of a resistor is equal to the product of the resistance and the current passing through the resistor. (iv) For the purpose of analysis, any combination of voltage sources and resistors can be replaced by a single voltage source and one resistor in series.
k k followed. Once again, this idea is almost intuitive. It can be illustrated with Figure 7.1.
The battery provides a voltage to force current through the resistors. All of the energy provided by the battery is spent in pushing current through the loop fromR1and thenR2 orR3. Furthermore, the energy spent going from point A to point B is the same whether the current goes throughR2orR3. The potential energy between A and B is also equal to the energy (voltage) of the battery minus the energy spent in passing current through the resistorR1.
In order to quantify voltages at various points in a circuit, probably the most useful rule is Ohm’s law that states the energy spent in pushing current through a resistor is the product of the current times the resistance.
Vresistor=iR (7.1)
The calculation of the voltage at point A with respect to the voltage of the ground and currents passing in different parts of the circuit in Figure 7.1 provides a good review of Ohm’s law. First of all, one can simplify the diagram by applying Thevenin’s theorem to replaceR2andR3with an effective resistor with resistanceR*. These two resistors provide parallel paths for current to flow between points A and B. The effective resistance, in ohms, of two resistors,R2andR3, in parallel can be shown to be equal to
R∗= R2⋅R3
R2+R3 = (2000)(3000)
2000+3000 = 6 000 000
5000 =1200 (7.2)
This expression can be derived using Kirchhoff’s laws and Ohm’s law. The equivalent circuit is shown in Figure 7.2.
The current,i1, that goes throughR1also goes throughR*. The energy spent in pushing this current through these resistors is given by Ohm’s law:
VR1 =i1R1 (7.3)
VR∗=i1R∗ (7.4)
1000 Ω
R* 1200 Ω
Ground 0 V
A R1
B Vbat
+5 V
FIGURE 7.2 The equivalent circuit for Figure 7.1 whereR* represents the equivalent resistance for the parallel resistorsR2andR3.
k k The sum of the voltages through the resistors must be equal to the voltage produced
by the battery.
Vbat=5.0 V=VR1+VR∗=i1R1+i1R∗=i1(R1+R∗) =i1(1000+1200) (7.5) Rearranging gives the current through the first resistor.
i1= Vbat
R1+R∗ = 5.0 V
2200Ω =2.27×10−3A (7.6)
Now applying Ohm’s law provides the voltage drop across resistor,R1:
VR1 =i1R1= (2.27×10−3A)(1000Ω) =2.27 V (7.7) The voltageVA, is also the voltage dropped across the equivalent resistor,R*.
VA=VR∗=i1R∗=
( Vbat R1+R∗
)
R∗= (2.27×10−3A)(1200) =2.72 V (7.8) It is worthwhile noting that the voltage,VA, is a fraction of the battery voltage,Vbat. The fraction is equal to the ratio ofR* and the total resistance,R1+R*. That is, the fraction of the total resistance between the battery and the ground represented byR*is equal to the fraction of the battery voltage that appears between point A and ground. Using two resis- tors in series such as shown in Figure 7.3a is a frequent strategy for dividing a voltage in a predictable way. In fact, this part of the circuit is called a voltage divider. A valuable way of implementing a voltage divider is to use a variable resistor, also known as a potentiometer, in place of the two resistors with fixed values. The potentiometer shown in Figure 7.3b has
Ground 0 V
(a) (b)
A R1
R2 B Vbat
+5 V R1 A
R2 Vbat
+5 V
FIGURE 7.3 (a) A voltage divider made from two resistors provides a voltage at point A that is a fraction of the supply voltage,Vbat. (b) A voltage divider based on a variable resistor. The arrow represents a moveable contact that can be placed anywhere along the length of the device to create complementary resistances,R1andR2, in order to provide any voltage betweenVbatand 0.0 at the moveable contact.
k k a movable contact represented by the arrow that can be positioned anywhere along the
path between the other two leads. The resistance between the moveable contact and the two separate ends defines two resistances,R1andR2, in the voltage divider. Consequently, applying a voltage,Vbat, to one end of the potentiometer and connecting the other end to ground permits one to select any fraction of the voltageVbat at the connection to the moveable contact (point A).
EXAMPLE 7.1
For the circuit in Figure 7.3a, choose resistors that will provide a voltage at point A of 1.00 V.
Equation (7.8) indicates that the voltage at point A compared to the ground (at point B) is equal to the battery voltage times the fraction of the total resistance (R1+R2) repre- sented by the resistance between A and B. That is,
VA= (5 V) ( R2
R1+R2 )
=1.00 or ( R2
R1+R2 )
=0.2
Lots of choices would work to give this ratio. A couple of practical considerations are helpful here. First of all, the choices for resistors above 1 MΩ are limited. Second, one might keep the total resistance at or above 1000Ωin order to keep the current level small (that is at or below a few milliamps) in order to keep from draining the battery.
That means the resistors should be in the range of 103to 106Ω. LetR2=10 000Ω. Then, ( 10 000
R1+10 000 )
=0.2 or R1=10 000
0.2 −10 000=40 000Ω