In the following circuit with an ideal operational amplifier, the capacitance of the parallel plate capacitor \(C\) is given by the expression \(C=\left(\dfrac{\varepsilon A}{x}\right)\), where \(\varepsilon\) is the dielectric constant of the medium between the capacitor plates, and \(A\) is the cross-sectional area. In the above relation, \(x\) is the separation between the two parallel plates, given by \(x=x_0+kt\), where \(t\) is time; \(x_0\) and \(k\) are positive non-zero constants. If the input voltage \(v_i\) is constant, then the output voltage \(v_0\) is given by __________________.
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For capacitors with time-varying capacitance \(C(t)\), use \(i = C\,\dfrac{dv}{dt} + v\,\dfrac{dC}{dt}\). If the voltage is constant, only the \(v\,\dfrac{dC}{dt}\) term contributes.
Step 1: The non-inverting terminal is grounded, hence the inverting node is at virtual ground (\(v_- \approx 0\)). The capacitor is connected between the input source \(v_i\) and the inverting node. Thus the voltage across the capacitor is \(v_i-0=v_i\), which is constant.
Step 2: For a time-varying capacitor, the current through it is
\[
i_C = C\,\frac{d(v_i-0)}{dt} + (v_i-0)\,\frac{dC}{dt} = v_i\,\frac{dC}{dt}
\]
since \(v_i\) is constant.
Step 3: KCL at the inverting node gives
\[
i_C + \frac{v_- - v_0}{R} = 0 \Rightarrow v_0 = R\,i_C = R\,v_i\,\frac{dC}{dt}.
\]
Step 4: With \(C=\dfrac{\varepsilon A}{x}\) and \(x=x_0+kt\),
\[
\frac{dC}{dt} = -\frac{\varepsilon A}{x^2}\,\frac{dx}{dt}
= -\frac{\varepsilon A}{x^2}\,k = -\frac{kC}{x}.
\]
Hence
\[
v_0 = R v_i \frac{dC}{dt} = -\,\frac{R v_i C k}{x}.
\]
(The negative sign indicates an inversion due to the inverting configuration; the magnitude matches option (A).)
Therefore, \(v_0 = \dfrac{R v_i C k}{x}\) (up to sign).
