The current gain ( I_out/I_in ) the circuit with an ideal current amplifier given below is

The correct answer is C_c/C_ƒ.

Here is the step-by-step derivation and explanation of the current gain for the given circuit:

1. Understanding the Op-Amp Behavior under Negative Feedback
The given circuit uses an ideal operational amplifier (op-amp) with negative feedback provided by the capacitor C_ƒ. For an ideal op-amp in negative feedback:
- The input currents to both the inverting (-) and non-inverting (+) terminals are zero.
- The virtual ground concept applies because the non-inverting terminal is connected to ground. Therefore, the voltage at the inverting node (V_-) is:

$$V_{-}=V_{+}=0\text{ V}$$

2. Applying Kirchhoff's Current Law (KCL) at the Inverting Node
From the schematic, the input current source I_in points downwards from the inverting terminal to ground, indicating that the current I_in is leaving the inverting node.
Let I_ƒ be the current flowing from the op-amp's output node (V_out) through the feedback capacitor C_ƒ into the inverting node. According to KCL at the inverting node:

$$I_{ƒ}=I_{\text{in}}$$

Using the current-voltage relationship of a capacitor, the current flowing through C_ƒ from V_out to the virtual ground node (V_- = 0) is:

$$I_{ƒ}=C_{ƒ}\frac{d(V_{\text{out}}-V_{-})}{dt}=C_{ƒ}\frac{d{V}_{\text{out}}}{dt}$$

Equating this to I_in gives:

$$C_{ƒ}\frac{d{V}_{\text{out}}}{dt}=I_{\text{in}}$$

Rearranging the equation to solve for the rate of change of the output voltage:

$$\frac{d{V}_{\text{out}}}{dt}=\frac{I_{\text{in}}}{C_{ƒ}}$$

3. Deriving the Output Current (I_out)
The output terminal has a capacitor C_c in series connected to ground, and the output current I_out flows from the op-amp output node V_out through C_c to the ground.
The current through the coupling capacitor C_c is:

$$I_{\text{out}}=C_c\frac{d{V}_{\text{out}}}{dt}$$

4. Calculating the Current Gain
Substitute the expression for $\frac{d{V}_{\text{out}}}{dt}$ obtained in step 2 into the equation for I_out:

$$I_{\text{out}}=C_c\left(\frac{I_{\text{in}}}{C_{ƒ}}\right)$$

Therefore, the ratio of the output current to the input current is:

$$\frac{I_{\text{out}}}{I_{\text{in}}}=\frac{C_c}{C_{ƒ}}$$

Thus, the current gain of the circuit is C_c/C_ƒ.