Suppose there is a circuit consisting of only resistances and batteries. Suppose one is to double (or increase it to $n$-times) all voltages and all resistances. Show that currents are unaltered.

(N/A) Let the equivalent internal resistance of the given cells be $R_{eq}$,the equivalent voltage be $V_{eq}$,and the external resistance be $R$.
From Ohm's law,the initial current $I_{1}$ is given by:
$I_{1} = \frac{V_{eq}}{R_{eq} + R}$ $(1)$
When the values of all resistors and all batteries are made $n$ times their initial values,the new parameters become:
$V'_{eq} = n V_{eq}$,$R'_{eq} = n R_{eq}$,and $R' = n R$.
The current in the new (modified) circuit,$I_{2}$,is:
$I_{2} = \frac{V'_{eq}}{R'_{eq} + R'} = \frac{n V_{eq}}{n R_{eq} + n R} = \frac{n V_{eq}}{n(R_{eq} + R)} = \frac{V_{eq}}{R_{eq} + R}$ $(2)$
Comparing $(1)$ and $(2)$,
$I_{2} = I_{1}$
Thus,the currents remain unaltered.