Why does a higher post-synaptic cell resistance lead to a higher voltage change when current is applied?
If you consider the electrical synapse a simple series RC circuit, with the resistance in the synapse and the capacitor as the postsynaptic cell, then you are correct that the value of R only changes the charge time for the capacitor (from a DC source, but not it's final voltage (at t = infinity). 



($\tau = RC$)

On the other hand, this model is too simplistic for the electrical synapse because you also have losses (discharge) to the outside of the cells. Image from the book:



And a more realistic circuit is created with extra (parallel) resistors below for the extracellular discharge. Note what happens when the 2nd resistor (discharge of postsynaptic cell to inter-cellular space) is introduced in the circuit (switch closed): the voltage charge on capacitor drops significantly. 


You basically have to "beat" or minimize this voltage drop, and one way to do this is to have a lower series resistor (another is of course a higher isolation [i.e. resistor] from the intercellular space). When the synapse resistance is 10 times lower than in the previous circuit, the voltage drop on the synapse is much lower, and so the voltage on the capacitor is higher, even with the switch closed:



You can play with the circuit here; it's set for the first set of resistor values; you need to right-click on a resistor and pick "edit" to change it's value; I could have added another switch and resistor to achieve that change for the synapse... but I didn't want to complicate the circuit conceptually.