Balancing Power
2.2 Electrical Resistance (Ohm’s Law)
R
R v R
i
v iR v i
viR
2.2 Electrical Resistance (Ohm’s Law) 31
where
If we choose the second method, we must write
(2.2)
where ,i, and R are, as before, measured in volts, amperes, and ohms, respectively. The algebraic signs used in Eqs. 2.1 and 2.2 are a direct conse- quence of the passive sign convention, which we introduced in Chapter 1.
Equations 2.1 and 2.2 are known as Ohm’s lawafter Georg Simon Ohm, a German physicist who established its validity early in the nine- teenth century. Ohm’s law is the algebraic relationship between voltage and current for a resistor. In SI units, resistance is measured in ohms. The Greek letter omega ( ) is the standard symbol for an ohm. The circuit diagram symbol for an resistor is shown in Fig. 2.7.
Ohm’s law expresses the voltage as a function of the current. However, expressing the current as a function of the voltage also is convenient. Thus, from Eq. 2.1,
(2.3)
or, from Eq. 2.2,
(2.4)
The reciprocal of the resistance is referred to as conductance, is sym- bolized by the letter G, and is measured in siemens (S). Thus,
(2.5)
An resistor has a conductance value of 0.125 S. In much of the profes- sional literature, the unit used for conductance is the mho (ohm spelled back- ward), which is symbolized by an inverted omega ( ). Therefore we may also describe an resistor as having a conductance of 0.125 mho, ( ).
We use ideal resistors in circuit analysis to model the behavior of physical devices. Using the qualifier ideal reminds us that the resistor model makes several simplifying assumptions about the behavior of actual resistive devices. The most important of these simplifying assump- tions is that the resistance of the ideal resistor is constant and its value does not vary over time. Most actual resistive devices do not have constant resistance, and their resistance does vary over time. The ideal resistor model can be used to represent a physical device whose resistance doesn’t vary much from some constant value over the time period of interest in the circuit analysis. In this book we assume that the simplifying assump- tions about resistance devices are valid, and we thus use ideal resistors in circuit analysis.
We may calculate the power at the terminals of a resistor in several ways. The first approach is to use the defining equation and simply calculate
Æ 8 Æ
Æ 8 Æ
G = 1 R S.
i = - v R. i = v
R, 8 Æ
Æ v
v = -iR,
R = the resistance in ohms.
i = the current in amperes, v = the voltage in volts,
8
Figure 2.7The circuit symbol for an 8 Æresistor.
Power in a resistor in terms of current
the product of the terminal voltage and current. For the reference systems shown in Fig. 2.6, we write
(2.6) when and
(2.7) when .
A second method of expressing the power at the terminals of a resis- tor expresses power in terms of the current and the resistance.
Substituting Eq. 2.1 into Eq. 2.6, we obtain
so
(2.8)
Likewise, substituting Eq. 2.2 into Eq. 2.7, we have
(2.9) Equations 2.8 and 2.9 are identical and demonstrate clearly that, regard- less of voltage polarity and current direction, the power at the terminals of a resistor is positive. Therefore, a resistor absorbs power from the circuit.
A third method of expressing the power at the terminals of a resistor is in terms of the voltage and resistance. The expression is independent of the polarity references, so
(2.10)
Sometimes a resistor’s value will be expressed as a conductance rather than as a resistance. Using the relationship between resistance and con- ductance given in Eq. 2.5, we may also write Eqs. 2.9 and 2.10 in terms of the conductance, or
(2.11) (2.12) Equations 2.6–2.12 provide a variety of methods for calculating the power absorbed by a resistor. Each yields the same answer. In analyzing a circuit, look at the information provided and choose the power equation that uses that information directly.
Example 2.3 illustrates the application of Ohm’s law in conjunction with an ideal source and a resistor. Power calculations at the terminals of a resistor also are illustrated.
p = v2G.
p = i2 G , p = v2
R.
p = -vi = -(-i R)i = i2R.
p = i2 R.
p = vi = (i R)i v = -i R
p = -vi v = i R
p = vi
Power in a resistor in terms of voltage
2.2 Electrical Resistance (Ohm’s Law) 33
Example 2.3 Calculating Voltage, Current, and Power for a Simple Resistive Circuit
In each circuit in Fig. 2.8, either the value of or iis not known.
Figure 2.8The circuits for Example 2.3.
a) Calculate the values of and i.
b) Determine the power dissipated in each resistor.
Solution
a) The voltage in Fig. 2.8(a) is a drop in the direc- tion of the current in the resistor. Therefore,
va = (1)(8) = 8 V.
va
v 1 A
(a) (b)
1 A
8 50 V 0.2 S
va
ib
(c) (d)
20 50 V 25
vc
id
v The current in the resistor with a conductance of 0.2 S in Fig. 2.8(b) is in the direction of the voltage drop across the resistor. Thus
The voltage in Fig. 2.8(c) is a rise in the direc- tion of the current in the resistor. Hence
The current in the resistor in Fig. 2.8(d) is in the direction of the voltage rise across the resistor. Therefore
b) The power dissipated in each of the four resistors is
p25Æ = (50)2
25 = (-2)2(25) = 100 W.
p20Æ = (-20)2
20 = (1)2(20) = 20 W, p0.2S = (50)2(0.2) = 500 W,
p8Æ = (8)2
8 = (1)2(8) = 8 W, id =
-50
25 = -2 A.
25 Æ id
vc = -(1)(20) = -20 V.
vc
ib = (50)(0.2) = 10 A.
ib
Objective 2—Be able to state and use Ohm’s Law . . .
A S S E S S M E N T P R O B L E M S
2.3 For the circuit shown,
a) If kV and mA, find the value of Rand the power absorbed by the resistor.
b) If mA and the power delivered by the voltage source is 3 W, find ,R, and the power absorbed by the resistor.
c) If and the power absorbed by R is 480 mW, find and .
Answer: (a) , 5 W;
(b) 40 V, , 3 W;
(c) 40 mA, 12 V.
533.33 Æ 200 kÆ
R ig
vg
vg
ig
R = 300 Æ
vg
ig = 75
ig = 5 vg = 1
2.4 For the circuit shown,
a) If A and , find and
the power delivered by the current source.
b) If and the power delivered to the conductor is 9 W, find the conductance G and the source current .
c) If and the power delivered to the conductance is 8 W, find and .
Answer: (a) 10 V, 5 W;
(b) 40 mS, 0.6 A;
(c) 40 mA, 200 V.
ig vg G
vg
ig
G = 200 mS
ig
vg = 15 V
vg
G = 50 mS ig = 0.5
NOTE: Also try Chapter Problems 2.11 and 2.12.
Having introduced the general characteristics of ideal sources and resis- tors, we next show how to use these elements to build the circuit model of a practical system.