Input impedance of a transistor amplifier from an example
The current through a resistor is proportional to the voltage across it:

$$I = \frac{V}{R}$$

This can also be written in terms of the change of current and the change of voltage:

$$\Delta I = \frac{\Delta V}{R}$$

If the collecter end of R1 was connected directly to a fixed voltage source VC, the voltage across it would be Vin &#150; VC, and any changes to this voltage would be attributed only to Vin. Therefore, the current changes through it would be:

$$\Delta I = \frac{\Delta V_{in} - \Delta V_C}{R} = \frac{\Delta V_{in}}{R}$$

since &Delta;VC is zero. We could calculate the effective resistance as:

$$R_{eff} = \frac{\Delta V_{in}}{\Delta I} = \Delta V_{in}\cdot \frac{R}{\Delta V_{in}} = {R}$$

All of this is pretty obvious, but what if VC varies, and does so in proportion to Vin:

$$V_C = A_V \cdot V_{in}$$

Now we have to write:

$$\Delta V = \Delta V_{in} - \Delta V_C = \Delta V_{in} - A_V \cdot \Delta V_{in} = \Delta V_{in}(1 - A_V)$$

Therefore:

$$\Delta I = \frac{\Delta V_{in}(1 - A_V)}{R}$$

and:

$$R_{eff} = \frac{\Delta V_{in}}{\Delta I} = \Delta V_{in}\cdot \frac{R}{\Delta V_{in}(1 - A_V)} = \frac{R}{1 - A_V}$$

Keeping in mind that AV is a negative number (a common-emitter amplifier inverts the signal), this tells us that the effective resistance is the real resistance divided by the gain of the amplifier.

In other words, if Vin varies by a little bit, the far end of the resistor swings in the opposite direction by a much larger amount, causing the current to be much larger than it would otherwise be, which makes the resistor seem much smaller than it actually is.