Determining Output Resistance of Transresistance Amplifier
The first thing you can do to simplify things a little bit is, transform the dependent voltage source \$A\cdot V_D\$ into a current source using Source Transformation. The equivalent circuit will look as shown below:



simulate this circuit &ndash; Schematic created using CircuitLab
Now we need to find: $$R_{output} = \frac{V_{output}}{I_x}$$
Calculating the sum of the currents at \$V_{output}\$:
$${\frac{A\cdot V_d}{R_o}} + I_x - \frac{V_{output}}{R_o} - \frac{V_{output}}{R_i + R_2} = 0$$
Using a current divider:
$$-\frac{V_{output}}{R_i + R_2}\cdot R_i = V_d$$
Replacing \$V_D\$ in the first equation yields:
$$-\frac{V_{output}}{R_i + R_2}\cdot R_i \cdot \frac{A}{R_o} + I_x - \frac{V_{output}}{R_o} - \frac{V_{output}}{R_i + R_2} = 0$$
Rearranging:
$$V_{output} \left[ \frac{R_i}{R_i + R_2}\cdot \frac{A}{R_o} + \frac{1}{R_o} + \frac{1}{R_i + R_2} \right] = I_x$$
Thus:
$$\frac{V_{output}}{I_x} = \frac{1}{\left[ \frac{R_i}{R_i + R_2}\cdot \frac{A}{R_o} + \frac{1}{R_o} + \frac{1}{R_i + R_2}\right]}$$
Rearranging:
$$R_{output} = \frac{R_o}{\left[ \frac{R_i}{R_i + R_2}\cdot A + \frac{R_o}{R_i + R_2} + 1 \right]}$$
The input resistance \$R_i\$ is usually very high so as \$R_i \rightarrow \infty\$ the equation above can be written as:
$$\frac{R_o}{1 + A}$$
Which is the output resistance divided by the loop gain thanks to the feedback provided by \$R_2\$.