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

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online