A series $$L - R$$ circuit is connected to a battery of emf $$V$$. If the circuit is switched on at $$t = 0$$, then the time at which the energy stored in the inductor reaches $$\left(\frac{1}{n}\right)$$ times of its maximum value, is:

We have a series $$L\!-\!R$$ circuit connected to a battery of emf $$V$$. When the key is closed at $$t = 0$$, current starts building up through the inductor as well as the resistor.

For an $$L\!-\!R$$ circuit, the standard growth-of-current formula is first stated:

$$i(t)=\frac{V}{R}\left(1-e^{-Rt/L}\right).$$

Here $$i(t)$$ is the instantaneous current at time $$t$$, $$V$$ is the applied emf, $$R$$ is the resistance and $$L$$ is the inductance.

The energy stored in the magnetic field of the inductor at any instant is given by the usual expression

$$U(t)=\frac12\,L\,i^2(t).$$

When the current has reached its final steady value, the inductor stores its maximum energy. The steady (saturated) current is obtained by putting $$t\to\infty$$ in the current formula:

$$I_0 = \frac{V}{R}\left(1-e^{-\infty}\right)=\frac{V}{R}.$$

So the maximum magnetic energy is

$$U_{\max}= \frac12\,L\,I_0^2 =\frac12\,L\left(\frac{V}{R}\right)^2.$$

The problem states that the energy at some time $$t$$ is only $$\dfrac1n$$ of this maximum value. Therefore we write

$$U(t)=\frac1n\,U_{\max}.$$

Substituting the explicit expressions for both energies,

$$\frac12\,L\,i^2(t)=\frac1n\left[\frac12\,L\left(\frac{V}{R}\right)^2\right].$$

Canceling the common factors $$\dfrac12 L$$ on both sides, we get the algebraic relation between currents:

$$i^2(t)=\frac1n\left(\frac{V}{R}\right)^2.$$

Taking the square root (and noting that current is positive during growth),

$$i(t)=\frac{V}{R}\,\frac1{\sqrt{n}}.$$

Now we replace $$i(t)$$ by its functional form $$\dfrac{V}{R}\left(1-e^{-Rt/L}\right)$$ :

$$\frac{V}{R}\left(1-e^{-Rt/L}\right)=\frac{V}{R}\,\frac1{\sqrt{n}}.$$

Dividing both sides by the common factor $$\dfrac{V}{R}$$ gives a simple transcendental equation in the exponential:

$$1-e^{-Rt/L}=\frac1{\sqrt{n}}.$$

Rearranging to isolate the exponential term,

$$e^{-Rt/L}=1-\frac1{\sqrt{n}}=\frac{\sqrt{n}-1}{\sqrt{n}}.$$

To remove the exponential, we take natural logarithms on both sides. Remember that $$\ln(e^{x}) = x$$:

$$-\frac{Rt}{L}=\ln\!\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right).$$

Multiplying by $$-\dfrac{L}{R}$$ converts it to an explicit expression for time $$t$$:

$$t=-\frac{L}{R}\,\ln\!\left(\frac{\sqrt{n}-1}{\sqrt{n}}\right).$$

The minus sign can be absorbed by inverting the logarithmic argument, using the property $$-\ln(x)=\ln\!\left(\dfrac1x\right)$$. Hence

$$t=\frac{L}{R}\,\ln\!\left(\frac{\sqrt{n}}{\sqrt{n}-1}\right).$$

This matches exactly the expression given in Option A.

Hence, the correct answer is Option A.