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Step-by-Step Solution
Step 1: Identify the given circuit parameters
• Self-inductance of the coil, $L = 10\,\text{mH} = 10 \times 10^{-3}\,\text{H}$.
• Resistance of the coil, $r_1 = 0.1\,\Omega$.
• Internal resistance of the battery, $r_2 = 0.9\,\Omega$.
• Hence, the total resistance in the circuit is $R = r_1 + r_2 = 0.1 + 0.9 = 1.0\,\Omega$.
• We need to find the time $t$ at which the current reaches $80\%\,$ of its final (saturation) value.
Step 2: Write the expression for the current growth in an LR circuit
When a DC source of emf $\varepsilon$ is connected to an inductor $L$ in series with a total resistance $R$, the current $i(t)$ grows according to:
$$i(t) = I_{\text{saturation}} \Bigl(1 - e^{-\frac{t}{\tau}}\Bigr),$$
where $I_{\text{saturation}} = \dfrac{\varepsilon}{R}$ is the final (steady-state) current and $\tau = \dfrac{L}{R}$ is the time constant of the circuit.
Step 3: Express the condition for $80\%$ of the saturation current
We have:
$$i(t) = 0.8 \, I_{\text{saturation}}.$$
From the growth equation,
$$0.8\, I_{\text{saturation}} = I_{\text{saturation}} \Bigl(1 - e^{-\frac{t}{\tau}}\Bigr).$$
Dividing both sides by $I_{\text{saturation}}$:
$$0.8 = 1 - e^{-\frac{t}{\tau}}.$$
So,
$$e^{-\frac{t}{\tau}} = 1 - 0.8 = 0.2.$$
Taking natural logarithm on both sides:
$$-\frac{t}{\tau} = \ln{(0.2)} = \ln{\left(\frac{1}{5}\right)} = -\ln{5}.$$
Hence,
$$\frac{t}{\tau} = \ln{5} \quad \Rightarrow \quad t = \tau\, \ln{5}.$$
Step 4: Calculate the time constant and determine the required time
• Time constant:
$$\tau = \frac{L}{R} = \frac{10 \times 10^{-3}\,\text{H}}{1.0\,\Omega} = 10 \times 10^{-3}\,\text{s} = 0.01\,\text{s}.$$
• Given $\ln{5} \approx 1.6,$ so
$$t = 0.01 \,\text{s} \times 1.6 = 0.016\,\text{s}.$$
Step 5: State the final answer
Therefore, the time taken for the current to reach $80\%\,$ of its saturation value is 0.016 s.
