Explanation Given Information: The provided formula for resistance R is 1 R = 1 R 1 + 1 R 2 + 1 R 3 . Where, the value of resistances R 1 = 4560 Ω , R 2 = 9150 Ω and R = 1830 Ω . Formula Used: Steps to solve the reciprocals with the help of calculator: Step 1: Simplify the reciprocal in terms of unknown letter. Step 2: Substitute the value of the known letters in the formula. Step 3: Now use the calculator and follow the steps. Calculation: Consider the provided formula of R , 1 R = 1 R 1 + 1 R 2 + 1 R 3 Subtract 1 R 1 + 1 R 2 on both side of the equation. 1 R − ( 1 R 1 + 1 R 2 ) = 1 R 1 + 1 R 2 + 1 R 3 − ( 1 R 1 + 1 R 2 ) 1 R − 1 R 1 − 1 R 2 = 1 R 1 + 1 R 2 + 1 R 3 − 1 R 1 −

Student Solutions Manual For Ewen/nelson's Elementary Technical Mathematics, 11th

11th Edition

ISBN: 9781285199276

Author: Dale Ewen

Publisher: Cengage Learning

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Chapter 6.9, Problem 16E

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To Calculate: The value of $R3$ if $R1=4560 Ω$ , $R2=9150 Ω$ and $R=1830 Ω$ . Where, the relation between them is $1R=1R1+1R2+1R3$ .

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Chapter 6.9, Problem 15E

Chapter 6.9, Problem 17E

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Chapter 6 Solutions

Student Solutions Manual For Ewen/nelson's Elementary Technical Mathematics, 11th

Ch. 6.1 - Solve each equation and check: x+2=8 Ch. 6.1 - Prob. 2E Ch. 6.1 - Prob. 3E Ch. 6.1 - Solve each equation and check: 23n=6 Ch. 6.1 - Solve each equation and check: w712=3 Ch. 6.1 - Prob. 6E Ch. 6.1 - Solve each equation and check: x13=1.5 Ch. 6.1 - Solve each equation and check: n+12=5 Ch. 6.1 - Solve each equation and check: 3b=15.6 Ch. 6.1 - Solve each equation and check: y17=25

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The value of R 3 if R 1 = 4560 Ω , R 2 = 9150 Ω and R = 1830 Ω . Where, the relation between them is 1 R = 1 R 1 + 1 R 2 + 1 R 3 .

Solution Summary: The author explains how to solve the reciprocals with the help of a calculator.

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To Calculate: The value of $R3$ if $R1=4560 Ω$ , $R2=9150 Ω$ and $R=1830 Ω$ . Where, the relation between them is $1R=1R1+1R2+1R3$ .

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Chapter 6 Solutions