[Johndavid1995]

Devise the series or parallel combination of 2 capacitances from the
E6 series that gives the closest match to the capacitance 1.14 μF.
Indicate the percentage error between the desired value and the
value you obtain.
This is my working out

0.68μF in parallel with 0.47μF and 1.0μF in parallel with 0.15μF
are pretty good.

Of all the possibilities, the best we have found is:
1.5μF in series with 4.7μF gives C = 1.137μF.

The % difference between the desired value (1.14μF) and 1.137μF is
100 x (1.14 -1.137)/1.14 = 0.26%

have I missed anything or have I got it right.

[ebaines]

Your math is correct, and unlike your first attempt it uses only two values of capacitance, so meets the requirements of the problem. However, I found an arrangement that uses just two values of caps but more than one of each - is that legal? If you put five 4.7 μF caps in series you get 0.94μF, and if you put eleven 2.2 μF caps in series you get 0.2 μF, then put these two sets in parallel to get 0.94 + 0.2 = 1.14 μF. Bingo!

[smoothy]

Problem is this is all theoretical. In the real world Capacitors have a very wide tolerance range, the worst of all electronic components except for coils. They could all be on the low end... or they could all be on the high end... and in fact I've seen enough of them OUTSIDE the tolerance range a long time ago back when I was still doing R&D work that one can never be certain they are even within their specified tolerance range. And while individually measuring and cherry picking them is possible for a batch... its not economically practical because you will be stuck with all the rest that can''t be returned to your parts supplier. And the bigger expense is time....the labor cost of the person measuring them, because time is money.

Realistically though...depending on their exact application in the design...the tolerance variance may not even matter most of the time. RF circuits are going to be the most sensitive.