Okay, berke buddy, you asked for it!
In English: Let's assume we have two resistors, which we'll wire either in series or in parallel. The value of the first resistor is random (since it has some random error), and we'll call its value X. Let's say its nominal value is one ohm, and the amount of error is normally distributed around one ohm. The value of the second resistor is also random, but it's independent of the first. We'll call its value Y.
In math speak: Assume two random variables, X and Y, which are indpendent, distributed randomly with Gaussian distribution, with mean one and variances \sigma _X ^2 and \sigma _Y ^2, respectively.
<br />
\begin{gathered}<br />
X \sim \left( {1,\sigma _X ^2 } \right) \hfill \\<br />
Y \sim \left( {1,\sigma _Y ^2 } \right) \hfill \\ <br />
\end{gathered} <br />
For the series case:
The sum of the two resistances in series is Z = X + Y. Since X and Y are independent and Gaussian, the sum Z is Gaussian, with mean two and variance equal to the sum of the variances of X and Y.
In math speak:
<br />
\begin{gathered}<br />
Z = X + Y \hfill \\<br />
Z \sim \left( {2,\sigma _X ^2 + \sigma _Y ^2 } \right) \hfill \\<br />
Var\left( Z \right) = \sigma _X ^2 + \sigma _Y ^2 \hfill \\ <br />
\end{gathered} <br />
If the variance of X is, say, 0.05, and the variance of Y is also 0.05, then the variance of the two resistors in series is 0.05 + 0.05 = 0.10. Keep this number in your head!
For the parallel case:
<br />
Z = \frac{1}<br />
{{\frac{1}<br />
{X} + \frac{1}<br />
{Y}}} = \frac{{XY}}<br />
{{X + Y}}<br />
We want to find the variance of Z:
<br />
Var\left( Z \right) = E\left( {Z^2 } \right) - E\left( Z \right)^2 <br />
so we need to find the second moment and expected value of Z.
<br />
\begin{gathered}<br />
E\left( Z \right) = \frac{{E\left( {XY} \right)}}<br />
{{E\left( {X + Y} \right)}} \hfill \\<br />
= \frac{{E\left( X \right)E\left( Y \right)}}<br />
{{E\left( X \right) + E\left( Y \right)}} \hfill \\<br />
= \frac{1}<br />
{2} \hfill \\ <br />
\end{gathered} <br />
The expected value of Z is 1/2, which makes perfectly good sense. If you put two resistors in parallel, each with resistance one ohm, you expect to get a combined resistance of 1/2 ohm.
The second moment of Z takes a bit more care to calculate:
<br />
\begin{gathered}<br />
E\left( {Z^2 } \right) = \frac{{E\left( {X^2 Y^2 } \right)}}<br />
{{E\left( {X^2 + 2XY + Y^2 } \right)}} \hfill \\<br />
= \frac{{E\left( {X^2 } \right)E\left( {Y^2 } \right)}}<br />
{{E\left( {X^2 } \right) + 2E\left( X \right)E\left( Y \right) + E\left( {Y^2 } \right)}} \hfill \\<br />
= \frac{{\left( {1 + \sigma _X ^2 } \right)\left( {1 + \sigma _Y ^2 } \right)}}<br />
{{\left( {1 + \sigma _X ^2 } \right) + 2 + \left( {1 + \sigma _Y ^2 } \right)}} \hfill \\<br />
= \frac{{\left( {1 + \sigma _X ^2 } \right)\left( {1 + \sigma _Y ^2 } \right)}}<br />
{{4 + \sigma _X ^2 + \sigma _Y ^2 }} \hfill \\ <br />
\end{gathered} <br />
Now we can find the variance of Z:
<br />
\begin{gathered}<br />
Var\left( Z \right) = E\left( {Z^2 } \right) - E\left( Z \right)^2 \hfill \\<br />
= \frac{{\left( {1 + \sigma _X ^2 } \right)\left( {1 + \sigma _Y ^2 } \right)}}<br />
{{4 + \sigma _X ^2 + \sigma _Y ^2 }} - \frac{1}<br />
{4} \hfill \\ <br />
\end{gathered} <br />
Now, if we assume the variances of X and Y are each 0.05, like we assumed above, the variance of Z, the parallel combination of the two, is only 0.019. The variance of the parallel combination of resistors is about 80% smaller than the variance of the series combination of resistors.
Thus, you're right -- the parallel combination is less sensitive to variations in each resistor's resistance than the series combination.
(This is graduate-level statistical inference, so I don't expect Younglearner to be able to follow all of it -- but the conclusions, I hope, are of some value!)
- Warren
