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

Options

A. `0.324s`
B. `0.103`s
C. `0.02s`
D. `0.016`s

Answer

To solve the problem step by step, we will follow the principles of an LR circuit and the formula for the current in such a circuit.

### Step 1: Identify the given values
- Self-inductance of the coil, \( L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H} \)
- Resistance of the coil, \( R_{\text{coil}} = 0.1 \, \Omega \)
- Internal resistance of the battery, \( R_{\text{internal}} = 0.9 \, \Omega \)

### Step 2: Calculate the total resistance in the circuit
The total resistance \( R \) in the circuit is the sum of the resistance of the coil and the internal resistance of the battery:
\[ R = R_{\text{coil}} + R_{\text{internal}} = 0.1 \, \Omega + 0.9 \, \Omega = 1.0 \, \Omega \]

### Step 3: Determine the saturation current \( I_0 \)
The saturation current \( I_0 \) can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
where \( V \) is the voltage of the battery. However, since we are not given the voltage, we can proceed without its explicit value since it will cancel out later.

### Step 4: Write the equation for current in an LR circuit
The current \( I(t) \) in an LR circuit is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where \( \tau \) (the time constant) is defined as:
\[ \tau = \frac{L}{R} \]

### Step 5: Calculate the time constant \( \tau \)
Substituting the values of \( L \) and \( R \):
\[ \tau = \frac{10 \times 10^{-3} \, \text{H}}{1.0 \, \Omega} = 10 \times 10^{-3} \, \text{s} = 0.01 \, \text{s} \]

### Step 6: Set up the equation for 80% of the saturation current
We want to find the time \( t \) when the current reaches 80% of its saturation value:
\[ I(t) = 0.8 I_0 \]
Substituting into the equation:
\[ 0.8 I_0 = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 0.8 = 1 - e^{-\frac{t}{\tau}} \]

### Step 7: Solve for \( e^{-\frac{t}{\tau}} \)
Rearranging gives:
\[ e^{-\frac{t}{\tau}} = 1 - 0.8 = 0.2 \]

### Step 8: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ -\frac{t}{\tau} = \ln(0.2) \]
Since \( \ln(0.2) = -\ln(5) \) (because \( 0.2 = \frac{1}{5} \)):
\[ -\frac{t}{\tau} = -\ln(5) \implies \frac{t}{\tau} = \ln(5) \]

### Step 9: Solve for time \( t \)
Now, substituting \( \tau \):
\[ t = \tau \ln(5) = 0.01 \, \text{s} \times 1.6 = 0.016 \, \text{s} \]

### Final Answer
The time taken for the current to attain 80% of the saturation value is:
\[ t = 0.016 \, \text{s} \]

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