REVIEW-Meters & Wheatstone Bridge

REVIEW-Meters & Wheatstone Bridge - ENGR 210 REVIEW...

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THINGS YOU SHOULD KNOW ABOUT METERS AND WHEATSTONE BRIDGES (Lecture 14) 1. The resistance of a meter. 2. How to analyze meter circuits for voltage and current 3. How to compute the open circuit voltage of a Wheatstone bridge 4. How to compute the Thevenin equivalent of a Wheatstone bridge. PAGE SIMPLE CURRENT METER 2 MULTI-RANGE AMMETER 3 VOLTMETER 4 WHEATSTONE BRIDGE 6-10 ENGR 210 REVIEW Meters & Wheatstone Bridges Page 1
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ENGR 210 Homework 2 Solutions Page4 Problem 3 HINT: Check out Example 3.5 A shunt resistor and a 50mV, 1ma D’Arsonval movement are used to build a 5 A ammeter. A resistance of 0.02 ohms is placed across the terminals of the ammeter. What is the new full scale range of the ammeter? a) The 1 st step is to determine the R A , When the current at the terminals is 5A, 1mA would flow through the meter coil, the current flow the resistor in parallel is 5-0.001=4.999A. And because the voltage is 50mV, therefore, 0.05/ 4.999 0.01 A R == ± b) when another 0.02 ± resistor is placed across the terminals, then the equivalent resistance of the two resistors ' A R = 0.00667 ± The new current flowing through the resistor is 0.05/0.00667=7.5A, Therefore, the new full scale range of the ammeter is 7.5A. Most circuits problems can be solved in multiple ways. Another way of looking at the problem is to recognize that 5A is flowing into the ammeter which has 50mV across it as shown below. R A 0.02 ± 5A 50mV i 50 ± ?A The new 0.02 ± resistor will also have 50mV across it so by Ohm’s Law the new 0.02 ± resistor will have current i = 50 mV 0.02 ± = 0.05 V 0.02 ± = 2.5 amps of current through it. The current ?A entering the meter will then be (by KCL) the sum of the 5A through the 5A meter and the additional 2.5A through the 0.02
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REVIEW-Meters & Wheatstone Bridge - ENGR 210 REVIEW...

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