262a then v 262 rx i the direct method of measuring

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Unformatted text preview: ecting a voltmeter in parallel with it, as shown in Fig. 2.62(a). Then V (2.62) Rx = I The direct method of measuring resistance is to use an ohmmeter. An ohmmeter consists basically of a d’Arsonval movement, a variable resistor or potentiometer, and a battery, as shown in Fig. 2.62(b). Applying KVL to the circuit in Fig. 2.62(b) gives Rn = A E = (R + Rm + Rx )Im I or Rx E − (R + Rm ) (2.63) Rx = Im The resistor R is selected such that the meter gives a full-scale deflection, that is, Im = Ifs when Rx = 0. This implies that E = (R + Rm )Ifs Rx = Ifs − 1 (R + Rm ) Im V (a) (2.64) Substituting Eq. (2.64) into Eq. (2.63) leads to + V − Ohmmeter Im (2.65) As mentioned, the types of meters we have discussed are known as analog meters and are based on the d’Arsonval meter movement. Another type of meter, called a digital meter, is based on active circuit elements such as op amps. For example, a digital multimeter displays measurements of dc or ac voltage, current, and resistance as discrete numbers, instead of using a pointer deflection on a continuous scale as in an analog multimeter. Digital meters are what you would most likely use in a modern lab. However, the design of digital meters is beyond the scope of this book. R Rm Rx E (b) Figure 2.62 Two ways of measuring resistance: (a) using an ammeter and a voltmeter, (b) using an ohmmeter. EXAMPLE 2.17 | v v Following the voltmeter setup of Fig. 2.60, design a voltmeter for the following multiple ranges: (a) 0 –1 V (b) 0 –5 V (c) 0 –50 V (d) 0 –100 V Assume that the internal resistance Rm = 2 k and the full-scale current Ifs = 100 µA. Solution: We apply Eq. (2.60) and assume that R1 , R2 , R3 , and R4 correspond with ranges 0 –1 V, 0 –5 V, 0 –50 V, and 0 –100 V, respectively. (a) For range 0 –1 V, 1 − 2000 = 10,000 − 2000 = 8 k R1 = 100 × 10−6 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents 60 PART 1 DC Circuits (b) For range 0 –5 V, R2 = 5 − 2000 = 50,000 − 2000 = 48 k 100 × 10−6 (c) For range 0 –50 V, R3 = 50 − 2000 = 500,000 − 2000 = 498 k 100 × 10−6 (d) For range 0 –100 V, R4 = 100 V − 2000 = 1,000,000 − 2000 = 998 k 100 × 10−6 Note that the rati...
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This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.

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