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If two resistances of value R 1 ...

If two resistances of value ` R _ 1 = ( 2.0 pm 0 .1 ) Omega and R _ 2 = ( 12.3 pm 0.2 ) ` are connected in (i) parallel and (ii) series then find the error in the estimation of equivalent resistance.

Text Solution

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The correct Answer is:
` R _ ( "series" ) = R _ 1 + R _ 2 `
` R _ ( "parallel" ) = ( R _ 1 + R _ 2 ) /( R _ 1 + R _ 2 ) `
(i) Series combination :
` R _ 1 = (2.0 + 12.3 ) pm ( 0 .1 + 0.2 ) Omega `
` = ( 14.3 pm 0 .3 ) Omega `
(ii) Parallel combination,
` R _ P = (2 xx 12.3 )/(14.3 ) = 1.720 Omega `
` ( Delta R _ P) /(R _ P ) = ( Delta R _ 1 ) /( R _ 1 ) + ( Delta R _ 2 ) /( R _ 2 ) + ( Delta ( R _ 1 + R _ 2 )) /( ( R _ 1 + R _ 2 )) = ( 0 .1 ) /(2) + ( 0.2 ) /( 12.3 ) + (( 0 .1 + 0.2 ))/(14.3 ) `
` ( Delta R _ P ) /(R _ P ) = (0.08727) `
` therefore Delta R _ P = 0.08727 xx R _ P = 0.08727 xx 1.720 `
` = 0.150 `
` therefore R _ P = (1.720 pm 0.150 ) `
` therefore R _ p = ( 1.72 pm 0.150 ) `
` = (1.7 pm 0.1 ) Omega `
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