In DC circuits, current is steady in time. Thus, there is no inductance, even if an inductor is present. An actual inductor, however, has resistance. This typically is on the order of 10 Ω. Often, an inductor’s resistance in a DC circuit is neglected.
FIGURE 2.8
A battery and electric motor circuit.
The first elementary DC circuit to analyze is that of a DC electric motor in series with a battery, as shown in Figure 2.8. Examine the battery first. It has an internal resistance, Rbatt, and an open circuit voltage (potential
difference), Voc. Rbattis the resistance and Voc the potential difference that
would be measured across the terminals of the battery if it were isolated from the circuit by not being connected to it. However, when the battery is placed in the circuit and the circuit switch is closed such that current, Ia,
flows around the circuit, the situation for the battery changes. The measured potential difference across the battery now is less because current flows through the battery, effectively leading to a potential difference across Rbatt.
This yields
Vbatt= Voc− IaRbatt, (2.17)
in which Vbatt represents the closed-circuit potential difference across the
battery. Similarly, the DC motor has an internal resistance, Ra, which is
mainly across its armature. It also has an opposing potential difference, Em, when operating with the battery connected to it. To summarize, Voc is
measured across the battery terminals when the switch is open, and Vbattis
Now what is Emin terms of the known quantities? To answer this, apply
Kirchhoff’s second law around the circuit loop when the switch is closed. Starting from the battery’s anode and moving in the direction of the current around the loop, this gives
Voc− IaRbatt− Em− IaRa= 0, (2.18)
which immediately leads to
Em= Voc− Ia(Rbatt+ Ra). (2.19)
This equation reveals a simple fact: the relatively high battery and motor internal resistances lead to a decrease in the motor’s maximum potential difference. This consequently results in a decrease in the motor’s power output to a device, such as the propeller of a remotely piloted aircraft.
FIGURE 2.9 An electrical circuit.
Example Problem 2.3
Statement: For the electrical circuit shown in Figure 2.9, determine [a] the magni- tude of the current in the branch between nodes A and B and [b] the direction of that current.
Solution: Application of Kirchhoff’s second law to the left loop gives 5 V − (1 Ω)(I1
A) + 2 V − (1 Ω)(I3 A) = 0. Similar application to the right loop yields 2 V − (2
Ω)(I2A) + (1 Ω)(I3A) = 0. At node A, application of Kirchhoff’s first law implies I1
− I2− I3= 0. These three expressions can be solved to yield I1= 3.8 A, I2 = 0.6 A,
and I3 = 3.2 A. Because I3is positive, the direction shown in Figure 2.9, from node A
to node B, is correct.
The second elementary direct-current circuit is a Wheatstone bridge. The Wheatstone bridge is used in a variety of common instruments such
FIGURE 2.10
The Wheatstone bridge configuration.
as pressure transducers and hot-wire anemometers. Its circuit, shown in Figure 2.10, consists of four resistors (R1through R4), each two comprising
a pair (R1and R2; R3and R4) in series that is connected to the other pair
in parallel, and a voltage source, Ei, connected between the R1-R3and the
R2-R4nodes. The voltage output of the bridge, Eo, is measured between the
R1-R2 and the R3-R4nodes. Eo is measured by an ideal voltmeter with an
infinite input impedance such that no current flows through the voltmeter. An expression needs to be developed that relates the bridge’s output voltage to its input voltage and the four resistances. There are four un- knowns, I1through I4. This implies that four equations are needed to reach
the desired solution. Examination of the circuit reveals that there are four closed loops for which four equations can be written by applying Kirchhoff’s second law. The resulting four equations are
Ei= I1R1+ I2R2, (2.20)
Ei= I3R3+ I4R4, (2.21)
Eo= I4R4− I2R2, (2.22)
and
Eo= −I3R3+ I1R1. (2.23)
Kirchhoff’s first law leads to I1= I2and I3= I4, assuming no current flows
through the voltmeter. These two current relations can be used in Equations 2.20 and 2.21 to give
I1=
Ei
R1+ R2
and
I3= Ei
R3+ R4. (2.25)
These two expressions can be substituted into Equation 2.23, yielding the desired result Eo= Ei R 1 R1+ R2 − R3 R3+ R4 . (2.26)
Equation 2.26 leads to some interesting features of the Wheatstone bridge. When there is no voltage output from the bridge, the bridge is considered to be balanced even if there is an input voltage present. This immediately yields the balanced bridge equation
R1
R2
= R3 R4
. (2.27)
This condition can be exploited to use the bridge to determine an unknown resistance, say R1, by having two other resistances fixed, say R2and R3, and
varying R4 until the balanced bridge condition is achieved. This is called
the null method. This method is used to determine the resistance of a sensor which usually is located remotely from the remainder of the bridge. An example is the hot-wire sensor of an anemometry system used in the constant-current mode to measure local fluid temperature.
FIGURE 2.11
Cantilever beam with four strain gages.
The bridge can be used also in the deflection method to provide an output voltage that is proportional to a change in resistance. Assume that resistance R1is the resistance of a sensor, such as a fine wire or a strain gage.
The sensor is located remotely from the remainder of the bridge circuit in an environment in which the temperature increases from some initial state. Its resistance will change by an amount δR from R1 to R
0 1. Application of Equation 2.26 yields Eo= Ei " R01 R0 1+ R2 − R3 R3+ R4 # . (2.28)
Further, if all the resistances are initially the same, where R1= R2= R3=
R4= R, then Equation 2.28 becomes
Eo= Ei δR/R 4 + 2δR/R = Eif (δR/R). (2.29)
Thus, when using the null and deflection methods, the Wheatstone bridge can be utilized to determine a resistance or a change in resistance.
One practical use of the Wheatstone bridge is in a force measurement system. This system is comprised of a cantilever beam, rigidly supported on one end, that is instrumented with four strain gages, two on top and two on bottom, as shown in Figure 2.11. The strain gage is discussed further in Chapter 3. A typical strain gage is shown in Figure 3.3 of that chapter. The electric configuration is called a four-arm bridge. The system operates by applying a known force, F , near the end of the beam along its centerline and then measuring the output of the Wheatstone bridge formed by the four strain gages. When a load is applied in the direction shown in Figure 2.11, the beam will deflect downward, giving rise to a tensile strain, L, on
the top of the beam and a compressive strain, −L, on the bottom of the
beam. Because a strain gage’s resistance increases with strain, δR ∼ L, the
resistances of the two tensile strain gages will increase and those of the two compressive strain gages will decrease. In general, following the notation in Figure 2.11, for the applied load condition,
R01= R1+ δR1, (2.30)
R04= R4+ δR4, (2.31)
R02= R2− δR2, (2.32)
and
R03= R3− δR3. (2.33)
If all four gages are identical, where they are of the same pattern with R1 = R2 = R3 = R4 = R, the two tensile resistances will increase by δR
and the two compressive ones will decrease by δR. For this case, Equation 2.28 simplifies to
Eo= Ei(δR/R). (2.34)
For a cantilever beam shown in Figure 2.11, the strain along the length of the beam on its top side is proportional to the force applied at its end, F . Thus, L∼ F . If strain gages are aligned with this axis of strain, then δR ∼ L, as
discussed in Section 3.3.2. Thus, the voltage output of this system, Eo, is
linearly proportional to the applied force, F. Further, with this strain gage configuration, variational temperature and torsional effects are compensated for automatically. This is an inexpensive, simple yet elegant measurement system that can be calibrated and used to determine unknown forces. This
configuration is the basis of most force balances used for aerodynamic and mechanical force measurements.
Example Problem 2.4
Statement: Referring to Figure 2.10, if R1 = 1 Ω, R2 = 3 Ω, and R3 = 2 Ω,
determine [a] the value of R4such that the Wheatstone bridge is balanced, and [b] the
bridge’s output voltage under this condition.
Solution: Equation 2.27 specifies the relationship between resistances when the bridge is balanced. Thus, R4= R2R3/R1= (3)(2)/1 = 6 Ω. Because the bridge is bal-
anced, its output voltage is zero. This can be verified by substituting the four resistance values into Equation 2.26.