The dimension of quantity (L / RCV) is

The correct option/answer is [A⁻¹].

To find the dimension of the quantity $\frac{L}{RCV}$, let us analyze the dimensional formula of each individual component: inductance ($L$), resistance ($R$), capacitance ($C$), and voltage/potential difference ($V$). We will relate these quantities to current ($I$ or $A$) and time ($T$).

First, recall the characteristic time constants of basic electrical circuits:
1. For an inductive-resistive (LR) circuit, the time constant is ${\tau }_L=\frac{L}{R}$. Therefore, the dimension of $\frac{L}{R}$ is time ($T$).
$\left[\frac{L}{R}\right]=\left[T\right]$
2. For a resistive-capacitive (RC) circuit, the time constant is ${\tau }_C=RC$. Therefore, the dimension of $RC$ is also time ($T$).

Now, let us analyze the denominator, $RCV$:
We know that the electric charge $Q$ stored in a capacitor is given by the relation:
$Q=CV$
Thus, the dimension of the product $CV$ is the dimension of charge ($Q$). Since current ($I$, represented dimensionally as $A$) is charge per unit time, we have:
$Q=I\cdot t$
Therefore, the dimensional formula of $CV$ is:
$\left[CV\right]=\left[AT\right]$

Substituting this back into the denominator $RCV$:
$\left[RCV\right]=\left[R\right]\cdot \left[CV\right]=\left[R\right]\cdot \left[AT\right]$

Alternatively, we can group the terms in the expression $\frac{L}{RCV}$ as follows:
$\frac{L}{RCV}=\left(\frac{L}{R}\right)\cdot \frac{1}{CV}$

Now, we substitute the dimensions we found:
- The dimension of $\frac{L}{R}$ is $\left[T\right]$.
- The dimension of $CV$ is $\left[AT\right]$.

Substituting these dimensions into the expression:
$\left[\frac{L}{RCV}\right]=\frac{\left[T\right]}{\left[AT\right]}$

Canceling the time dimension $T$ from both the numerator and the denominator, we get:
$\left[\frac{L}{RCV}\right]=\frac{1}{\left[A\right]}=\left[A^{-1}\right]$

Thus, the dimension of the quantity is indeed $\left[A^{-1}\right]$.