Lets start with the EMF equation, $\epsilon = I(r+R)$, where $\epsilon$ is EMF, $R$ is the resistance of the circuit, $r$, internal resistance and finally $I$, current. If we are given the values, $I = 2 \pm 0.1$ $A$, $V = 1.5 \pm 0.2$ $Volts$ where $V$ is the terminal potential difference, and $\epsilon$ $= 4.5 \pm 0.2$ $Volts$. From the EMF equation we can expand and get $\epsilon = Ir + IR$, and finally $\epsilon = V + Ir$, where $V$ is terminal pd. Rearranging for $r$ gives $\frac{\epsilon - v}{I}$. The rules we have are, if $Z = A \pm B$, the absolute uncertainty in $Z$ is just the sum of the absolute uncertainties in $A$ and $B$, and if $Z = A *B$, the percentage uncertainty in $Z$ is the sum of the percentage uncertainties in $A$ and $B$, the same rule goes for $Z = \frac{A}{B}$. Putting this all together we know the percentage uncertainty in $r$ will be the sum of the percentage uncertainties in $\epsilon - V$ and $I$, giving us $\frac{100 *0.3}{1.5} + \frac{100*0.1}{2}$ $= 25$. This is where the confusion starts. What if I had rearranged it to $\epsilon - V = Ir$, we can find the percentage uncertainty in $\epsilon - V$ to be $20$ just like before, but we know that the percentage uncertainty in $\epsilon - V$ will be equal to that of $Ir$, so $20$ equals the percentage uncertainty in $r +$ the percentage uncertainty in $I$, but this gives $15$ percent, not $25$. Where is my misunderstanding?
1 Answer
Let $D=\epsilon-V$, so you have $D=Ir$. The uncertainty propagation is NOT symmetric. i.e. uncertainty of D given $I$ and $r$ uncertainties is different from the uncertainty of $r$ given uncertainty of $D$ and $I$.
The formulæ you are using right now are but conceptually simpler approximations. You will likely get the chance to learn about the later versions, that will be helpful in understanding this behaviour further.
In more precise terms, $$\frac{\Delta D}D=\frac{\Delta I}I+\frac{\Delta r}r\qquad\text{does NOT imply}\qquad\frac{\Delta r}r=\frac{\Delta D}D-\frac{\Delta I}I$$but rather the one with the positive sign instead of negative. Which equation to use, depends upon which uncertainty is unknown and which are, so that the propagation defines a direction, and that direction will be the one the determines which equation to use.
