Input Resistance: Why can you ignore current sources?
Say you have this circuit, and you apply some voltage to the terminals on the left:







simulate this circuit &ndash; Schematic created using CircuitLab

What's the input impedance? \$1\Omega\$, obviously. Why? Say we want to increase the current by \$1A\$. By how much will be have to increase the current? By Ohm's law:

$$ E = 1A\cdot 1\Omega = 1V$$

That is, we need to increase the voltage by \$1V\$ per \$1A\$. That is, one ohm is one volt per ampere:

$$ 1\Omega = \frac{1V}{1A} $$

Consider this circuit:







simulate this circuit

How much must voltage increase to increase current by \$1A\$? It takes a much larger increase in voltage to increase the current now:

$$ 1k\Omega = \frac{1000V}{1A} $$

What about this circuit?







simulate this circuit

What about this one?







simulate this circuit

By how much would you have to increase the voltage to increase the current? You can't. As the input impedance approaches infinity, the change in current per change in voltage becomes less. In the limiting case, the load is a current source that never changes, and input impedance is infinite.

Here's another thought experiment. Consider these circuits:







simulate this circuit

How does the current vary in each as the load is varied? As the series resistance (R1 or R2) increases, the current varies less with load. If \$R_{load} \ll R_2\$, then the current hardly changes at all. In the limiting case, when that resistance is infinite, current doesn't change at all, and you have a current source. Of course you can't pass any current through an infinite resistance, but then you can't have an ideal current source, either.