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

The correct answer is:

$\frac{C_c}{C_f}$

Circuit Diagram:

Step-by-Step Derivation:

1. Virtual Ground Concept:
The circuit consists of an ideal operational amplifier indicated by the infinity symbol ($\infty$). The non-inverting input terminal is connected directly to ground:
$V_{+}=0\text{V}$
Due to the negative feedback through the capacitor $C_f$, the virtual short condition dictates that the voltage at the inverting input node is also at ground potential:
$V_{-}=V_{+}=0\text{V}$

2. Nodal Analysis at the Inverting Input Node:
Since the operational amplifier is ideal, no current flows into its input terminals. Applying Kirchhoff's Current Law (KCL) at the inverting node:
$I_{\text{in}}=I_{C_f}$
where $I_{C_f}$ is the current flowing through the feedback capacitor $C_f$ from the output node (voltage $V_{\text{out}}$) to the inverting input node (voltage $V_{-}=0$):
$I_{C_f}=C_f\frac{d(V_{\text{out}}-V_{-})}{dt}=C_f\frac{d{V}_{\text{out}}}{dt}$
Therefore, we obtain the relation for the input current:
$I_{\text{in}}=C_f\frac{d{V}_{\text{out}}}{dt}$

3. Analyzing the Output Current:
The output current $I_{\text{out}}$ is the current flowing from the output node of the op-amp through the capacitor $C_c$ directly to ground:
$I_{\text{out}}=C_c\frac{d{V}_{\text{out}}}{dt}$

4. Calculating the Current Gain:
By substituting the expression for the rate of change of output voltage $\frac{d{V}_{\text{out}}}{dt}=\frac{I_{\text{in}}}{C_f}$ into the equation for $I_{\text{out}}$, we get:
$I_{\text{out}}=C_c\left(\frac{I_{\text{in}}}{C_f}\right)$
Dividing both sides by $I_{\text{in}}$ yields the overall current gain:
$\frac{I_{\text{out}}}{I_{\text{in}}}=\frac{C_c}{C_f}$