4.10 Applications

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4.10 Applications

In this section we will discuss tw o important practical applications of the concepts co vered in this chapter: source modeling and resistance measurement.

4.10.1 Source Modeling

Source modeling provides an example of the usefulness of the Thevenin or the Norton equivalent. An active source such as a battery is often characterized by its Thevenin or Norton equivalent circuit. An ideal voltage source provides a constant v oltage irrespective of the current dra wn by the load, while an ideal current source supplies a constant current regardless of the load voltage. As Fig. 4.58 shows, practical voltage and current sources are not ideal, due to their internal resistances or source resistances Rs and Rp. They become ideal as Rs → 0 and Rp → ∞. To show that this is the case, consider the effect of the load on voltage sources,

154 Chapter 4 Circuit Theorems

as shown in Fig. 4.59(a). By the voltage division principle, the load voltage is

$$v_L = \frac{R_L}{R_s + R_L} v_s \tag{4.25}$$

As RL increases, the load v oltage approaches a source v oltage vs, as illustrated in Fig. 4.59(b). From Eq. (4.25), we should note that:

1. The load v oltage will be constant if the internal resistance Rs of the source is zero or , at least, Rs ≪ RL. In other w ords, the smaller Rs is compared with RL, the closer the v oltage source is to being ideal.

(a) Practical voltage source connected to a load RL, (b) load voltage decreases as RL decreases.

2. When the load is disconnected (i.e., the source is open-circuited so that RL → ∞), voc = vs. Thus, vs may be re garded as the unloaded source voltage. The connection of the load causes the terminal voltage to drop in magnitude; this is known as the loading effect.

The same argument can be made for a practical current source when connected to a load as sho wn in Fig. 4.60(a). By the current di vision principle,

$$i_L = \frac{R_p}{R_p + R_L} i_s \tag{4.26}$$

Figure 4.60(b) shows the variation in the load current as the load resis tance increases. Again, we notice a drop in current due to the load (loading effect), and load current is constant (ideal current source) when the internal resistance is very large (i.e., Rp → ∞ or, at least, Rp ≫ RL).

Sometimes, we need to know the unloaded source voltage vs and the internal resistance Rs of a v oltage source. To find vs and Rs, we follo w the procedure illustrated in Fig. 4.61. First, we measure the open-circuit voltage voc as in Fig. 4.61(a) and set

$$v_s = v_{oc} \tag{4.27}$$

Then, we connect a v ariable load RL across the terminals as in Fig. 4.61(b). We adjust the resistance RL until we measure a load voltage of exactly one-half of the open-circuit voltage, v^L = voc∕2, be cause now RL = RTh = Rs. At that point, we disconnect RL and measure it. We set

$$R_s = R_L \tag{4.28}$$

For example, a car battery may have v^s = 12 V and Rs = 0.05 Ω.

Figure 4.60

(a) Practical current source connected to a load RL, (b) load current decreases as RL increases.

(a) Measuring voc, (b) measuring vL.

The terminal v oltage of a v oltage source is 12 V when connected to a Example 4.16 2-W load. When the load is disconnected, the terminal v oltage rises to 12.4 V. (a) Calculate the source v oltage vs and internal resistance Rs. (b) Determine the voltage when an 8-Ω load is connected to the source.

Solution:

(a) We replace the source by its Thevenin equivalent. The terminal voltage when the load is disconnected is the open-circuit voltage,

$$v_s = v_{oc} = 12.4 \text{ V}$$

When the load is connected, as shown in Fig. 4.62(a), v^L = 12 V and p^L = 2 W. Hence,

$$p_L = \frac{v_L^2}{R_L}$$

$\Rightarrow$ $R_L = \frac{v_L^2}{p_L} = \frac{12^2}{2} = 72 \Omega$

The load current is

$$i_L = \frac{v_L}{R_L} = \frac{12}{72} = \frac{1}{6}$$

A

The voltage across Rs is the difference between the source voltage vs and the load voltage vL, or

$$12.4 - 12 = 0.4 = R_s i_L, \qquad R_s = \frac{0.4}{I_L} = 2.4 \ \Omega$$

(b) Now that we have the Thevenin equivalent of the source, we connect the 8-Ω load across the Thevenin equivalent as shown in Fig. 4.62(b). Using voltage division, we obtain

$$v = \frac{8}{8 + 2.4} (12.4) = 9.538 \text{ V}$$

Figure 4.62 For Example 4.16.

The measured open-circuit voltage across a certain amplifier is 16 V. The voltage drops to 8 V when a 8-Ω loudspeaker is connected to the amplifier. Calculate the voltage when a 24-Ω loudspeaker is used instead.

Practice Problem 4.16

Answer: 12 V.

4.10.2 Resistance Measurement

Although the ohmmeter method pro vides the simplest w ay to mea sure resistance, more accurate measurement may be obtained using the Wheatstone bridge. While ohmmeters are designed to measure resistance in low, mid, or high range, a Wheatstone bridge is used to mea sure resistance in the mid range, say, between 1 Ω and 1 MΩ. Very low values of resistances are measured with a milliohmmeter, while very high values are measured with a Megger tester.

The Wheatstone bridge (or resistance bridge) circuit is used in a number of applications. Here we will use it to measure an unkno wn resistance. The unkno wn resistance Rx is connected to the bridge as shown in Fig. 4.63. The variable resistance is adjusted until no current flows through the galvanometer, which is essentially a d’Arsonval movement operating as a sensiti ve current-indicating device like an ammeter in the microamp range. Under this condition v¹ = v2, and the bridge is said to be balanced. Since no current flows through the galvanometer, R¹ and R2 behave as though the y were in series; so do R3 and Rx. The fact that no current flows through the galvanometer also implies that v¹ = v2. Applying the voltage division principle,

$$v_1 = \frac{R_2}{R_1 + R_2} v = v_2 = \frac{R_x}{R_3 + R_x} v$$

\n(4.29)

Hence, no current flows through the galvanometer when

$$\frac{R_2}{R_1 + R_2} = \frac{R_x}{R_3 + R_x} \qquad \Rightarrow \qquad R_2 R_3 = R_1 R_x$$ $$R_x = \frac{R_3}{R_1} R_2 \tag{4.30}$$

If R¹ = R3, and R2 is adjusted until no current flows through the galvanometer, then Rx = R2.

How do we find the current through the galvanometer when the Wheatstone bridge is unbalanced? We find the Thevenin equivalent (VTh and RTh) with respect to the g alvanometer terminals. If Rm is the resis tance of the g alvanometer, the current through it under the unbalanced condition is

$$I = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_m} \tag{4.31}$$

Example 4.18 will illustrate this.

In Fig. 4.63, R¹ = 500 Ω and R³ = 200 Ω. The bridge is balanced when R² is adjusted to be 125 Ω. Determine the unknown resistance Rx.

Solution:

or

Using Eq. (4.30) gives

$$R_x = \frac{R_3}{R_1} R_2 = \frac{200}{500} 125 = 50 \ \Omega$$

Historical note: The bridge was invented by Charles Wheatstone (1802–1875), a British professor who also invented the telegraph, as Samuel Morse did independently in the

United States.

Figure 4.63 The Wheatstone bridge; Rx is the resistance to be measured.

A Wheatstone bridge has R¹ = R³ = 2 kΩ. R2 is adjusted until no current flows through the galvanometer. At that point, R² = 6.3 kΩ. What is the value of the unknown resistance?

Answer: 6.3 kΩ.

The circuit in Fig. 4.64 represents an unbalanced bridge. If the galvanometer has a resistance of 40 Ω, find the current through the galvanometer.

Unbalanced bridge of Example 4.18.

Solution:

We first need to replace the circuit by its Thevenin equivalent at ter minals a and b. The Thevenin resistance is found using the circuit in Fig. 4.65(a). Notice that the 3-kΩ and 1-KΩ resistors are in parallel; so are the 400 and 600- Ω resistors. The two parallel combinations form a series combination with respect to terminals a and b. Hence,

$$R_{\text{Th}} = 3000 \parallel 1000 + 400 \parallel 600$$

=

$$\frac{3000 \times 1000}{3000 + 1000} + \frac{400 \times 600}{400 + 600} = 750 + 240 = 990 \Omega$$

To find the Thevenin voltage, we consider the circuit in Fig. 4.65(b). Using the voltage division principle gives

sing the voltage division principle gives

$$v_1 = \frac{1000}{1000 + 3000} (220) = 55 \text{ V}, \qquad v_2 = \frac{600}{600 + 400} (220) = 132 \text{ V}$$

Applying KVL around loop ab gives

$$-v_1 + V_{\text{Th}} + v_2 = 0$$

or $V_{\text{Th}} = v_1 - v_2 = 55 - 132 = -77$ V

Having determined the Thevenin equivalent, we find the current through the galvanometer using Fig. 4.65(c).

$$I_G = \frac{V_{\text{Th}}}{R_{\text{Th}} + R_m} = \frac{-77}{990 + 40} = -74.76 \text{ mA}$$

The negative sign indicates that the current flows in the direction oppo site to the one assumed, that is, from terminal b to terminal a.

Practice Problem 4.17

Example 4.18

For Example 4.18: (a) Finding RTh, (b) finding VTh, (c) determining the current through the galvanometer.

Practice Problem 4.18 14 Ω 60 Ω 16 V 40 Ω 20 Ω 30 Ω G

Figure 4.66 For Practice Prob. 4.18.

Obtain the current through the g alvanometer, ha ving a resistance of 14 Ω, in the Wheatstone bridge shown in Fig. 4.66.

Answer: 64 mA.