Ignore cite evidence Discussion How does the % error compare to the coded tolera

Question

Ignore cite evidence Discussion How does the % error compare to the coded tolerance for your resistors? What is the apparent rule for combining equal resistances in series circuits? In parallel circuits? Cite evidence from your data to support your conclusions. circuits? Cite evidence from your data to support your conclusions. parallel? Cite evidence from your data to support your conclusions what is the apparent rule for combining unequal resistances in senes circuits? In parallel What is the apparent rule for the total resistance when resistors are added up in series? In

Explanation / Answer

1. Tolerance is the % error in the value of a resistance. So %error in value of a single resistance is equal to its coded tolerance

2. for combining equal resistances in series, the total resistance becomes twise the given resistnace

the total absolute error is added algebriacally

hence if the resistance was R and tolerance was k% then

absolute error in R = dR = Rk/100

hence in series addition

net error in total resistance, dR' = 2Rk/100

net reissitance R' = 2R

hence

net tolerance = 2Rk/100*2R = k%

hence in series addition of equal resistnaces the tolerance remains the same

3. for combining unequal resistnaces in series

let resistnaces be R and r

let tolerances be k% and m%

then absolute errors are dR = Rk/100, dr = rm/100

so in series addition

net error dR' = (Rk + rm)/100

net reisstance, R' = r + R

hence net tolerance = dR'/R' = (Rk + rm)/(R + r)100

so if the unequal resitances have same tolerance k

dR'/R' = k% [ tolerance of overall resistnace remains the same as original resistance tolerance ]

for unequal tolerances

dR'/R' = (Rk + rm)/(R + r) %

4. rule for adding total resistance in series = adding algebriacally

let there be two resistnaces r and R

then in series addition Reff = r + R

in parallel addition

Reff = rR/(R + r)

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