A coil of self inductance 10 mH and resistance of 0.1 Ω is connected through a switch to a battery of internal resistance 0.9 Ω. After the switch is closed, the time taken for the current to attain 80% of the saturation value is: [ln5 = 1.6]

We have a circuit in which a coil of self-inductance $$L = 10\,\text{mH} = 10 \times 10^{-3}\,\text{H}$$ and resistance $$R_{\text{coil}} = 0.1\,\Omega$$ is connected to a battery whose internal resistance is $$r = 0.9\,\Omega$$. When the switch is closed, the coil resistance and the internal resistance are in series, so the total series resistance is

$$R = R_{\text{coil}} + r = 0.1\,\Omega + 0.9\,\Omega = 1.0\,\Omega.$$

The growth of current in an $$L\!-\!R$$ circuit after the switch is closed is governed by the exponential law

$$I(t) = I_0 \left(1 - e^{-t/\tau}\right),$$

where $$I_0$$ is the steady (saturation) current and $$\tau$$ is the time constant. The time constant for an $$L\!-\!R$$ circuit is given by the formula

$$\tau = \frac{L}{R}.$$

Substituting the given values, we obtain

$$\tau = \frac{10 \times 10^{-3}\,\text{H}}{1.0\,\Omega} = 10^{-2}\,\text{s} = 0.01\,\text{s}.$$

The problem asks for the time $$t$$ required for the current to reach 80 % of its saturation value. Mathematically, this means

$$I(t) = 0.8\,I_0.$$

Using the current-growth equation, we write

$$0.8\,I_0 = I_0 \left(1 - e^{-t/\tau}\right).$$

Dividing both sides by $$I_0$$, we get

$$0.8 = 1 - e^{-t/\tau}.$$

Rearranging to isolate the exponential term,

$$e^{-t/\tau} = 1 - 0.8 = 0.2.$$

Taking natural logarithms on both sides, we have

$$-\,\frac{t}{\tau} = \ln(0.2).$$

Now, $$0.2 = \frac{1}{5},$$ so

$$\ln(0.2) = \ln\!\left(\frac{1}{5}\right) = -\ln 5.$$

The question supplies $$\ln 5 = 1.6,$$ hence

$$\ln(0.2) = -1.6.$$

Substituting this value, we find

$$-\,\frac{t}{\tau} = -1.6 \quad\Longrightarrow\quad \frac{t}{\tau} = 1.6.$$

Finally, substituting $$\tau = 0.01\,\text{s}$$, we get

$$t = 1.6 \times 0.01\,\text{s} = 0.016\,\text{s}.$$

Hence, the correct answer is Option D.