Time Constant of Discharging Parallel Plate Capacitor with Decreasing Liquid Level

To find the time constant of the given circuit as a function of time $t$, we need to determine the equivalent capacitance $C(t)$ of the capacitor at any time $t$.

1. Analysis of the Capacitor Geometry

• The area of the capacitor plates is $A = 1$.
• The distance between the parallel plates is $d$.
• The liquid level decreases at a constant speed $V$.

At time $t$, the thickness of the liquid layer $x(t)$ is given by:
$x(t) = \frac{d}{3} - Vt$

Consequently, the thickness of the air medium above the liquid is:
$d - x(t) = d - \left(\frac{d}{3} - Vt\right) = \frac{2d}{3} + Vt$

2. Equivalent Capacitance Calculation

The liquid medium and the air medium form two dielectric layers oriented parallel to the capacitor plates, which can be modeled as two capacitors connected in series:

1. Capacitance of the liquid layer ($C_1$):
   $C_1 = \frac{K \varepsilon_0 A}{x(t)} = \frac{2 \varepsilon_0 (1)}{\frac{d}{3} - Vt}$

2. Capacitance of the air layer ($C_2$):
   $C_2 = \frac{\varepsilon_0 A}{d - x(t)} = \frac{\varepsilon_0 (1)}{\frac{2d}{3} + Vt}$

Using the formula for series combination of capacitors, $\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2}$:
$\frac{1}{C(t)} = \frac{\frac{d}{3} - Vt}{2\varepsilon_0} + \frac{\frac{2d}{3} + Vt}{\varepsilon_0}$

Taking a common denominator $2\varepsilon_0$:
$\frac{1}{C(t)} = \frac{\left(\frac{d}{3} - Vt\right) + 2\left(\frac{2d}{3} + Vt\right)}{2\varepsilon_0}$
$\frac{1}{C(t)} = \frac{\frac{d}{3} - Vt + \frac{4d}{3} + 2Vt}{2\varepsilon_0} = \frac{\frac{5d}{3} + Vt}{2\varepsilon_0}$
$\frac{1}{C(t)} = \frac{5d + 3Vt}{6\varepsilon_0}$

Inverting this expression gives the equivalent capacitance $C(t)$:
$C(t) = \frac{6\varepsilon_0}{5d + 3Vt}$

3. Time Constant ($\tau$)

, The time constant $\tau(t)$ of an $RC$ circuit is defined as:
$\tau(t) = R C(t) = \frac{6\varepsilon_0 R}{5d + 3Vt}$

Thus, the correct option is A.