Question

Doubts:

Answer 1

4 1 ohm resistances of tolerance 5% each are connected in series. Find the tolerance of equivalent resistance?

Answer 2

@Vincent-Lyon.Fr

Answer 3

for resitances in series:
Tolerance = \[\frac{ \sum_{}^{}RT }{ \sum_{}^{}R }\]

Answer 4

Sorry, I'm not an expert in this field, but I think

\(R_{eq}=R_1+R_2+R_3\)

\(\dfrac{\Delta R_{eq}}{R_{eq}}=\dfrac{\Delta (R_1+R_2+R_3)}{R_1+R_2+R_3}=\dfrac{3\Delta R_1}{3R_1}=\dfrac{\Delta R_1}{R_1}=5\)%

Answer 5

An interesting approach to this problem can be found through a statistical analysis which assumes a normal distribution of the values (although in practice the distribution might not be normal). Such an analysis can be found here:
http://paulorenato.com/joomla/index.php?option=com_content&view=article&id=109&Itemid=4
Given n resistors of mean value
\[E(R _{i})=\mu\]
and tolerance
\[T _{i}\]

Answer 6

the tolerance of the resulting resistor Tt is
\[\frac{T _{i}}{\sqrt{n}}\]

Answer 7

So taking the values in the question we get the initial tolerance of 5% for each resistor reduced to
\[\frac{5}{\sqrt{4}}=2.5\ percent\]
for the resulting total resistance.

Answer 8

So at the worst (with the most poorly designed resistors within the toloerance limits) the tolerance will be 5%
though NORMALIZED tolerance is 2.5%

Answer 9

The options given were 5%,10%,15%,20%. I guess I will go with Vince's method.

Answer 10

If the values followed a continuous uniform (rectangular) distribution the tolerance of the equivalent resistance would indeed be 5%. However such a distribution would clearly show that the manufacturer selected the resistances to meet the specified tolerance rather than manufactured them to meet a specific mean value by means of a well controlled process.

Answer 11

Thanks @kropot72 :)

Answer 12

You're most welcome :)