A coil having inductance and L and resistance R is connected to a battery of emf `in` at `t = 0`. If `t_(1)` and `t_(2)` are time for `90%` and `99%` completion of current growth in the circuit, then `(t_(1))/(t_(2))` will be-

Answer

To solve the problem, we need to find the ratio \( \frac{t_1}{t_2} \), where \( t_1 \) is the time taken for the current to reach 90% of its maximum value and \( t_2 \) is the time taken for the current to reach 99% of its maximum value in an RL circuit.

### Step-by-Step Solution:

1. **Understanding the RL Circuit**:
In an RL circuit, the current \( I(t) \) at time \( t \) when connected to a battery of emf \( E \) is given by:
\[ I(t) = \frac{E}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
where \( R \) is the resistance and \( L \) is the inductance.

2. **Finding the Maximum Current**:
The maximum current \( I_{max} \) in the circuit is:
\[ I_{max} = \frac{E}{R} \]

3. **Setting Up for 90% Completion**:
For \( t_1 \), we want the current to reach 90% of \( I_{max} \):
\[ I(t_1) = 0.9 I_{max} = 0.9 \frac{E}{R} \]
Substituting into the equation for \( I(t) \):
\[ 0.9 \frac{E}{R} = \frac{E}{R} \left(1 - e^{-\frac{R}{L}t_1}\right) \]
Dividing both sides by \( \frac{E}{R} \):
\[ 0.9 = 1 - e^{-\frac{R}{L}t_1} \]
Rearranging gives:
\[ e^{-\frac{R}{L}t_1} = 0.1 \]

4. **Taking Natural Logarithm**:
Taking the natural logarithm on both sides:
\[ -\frac{R}{L}t_1 = \ln(0.1) \]
Thus, we find:
\[ t_1 = -\frac{L}{R} \ln(0.1) \]

5. **Setting Up for 99% Completion**:
For \( t_2 \), we want the current to reach 99% of \( I_{max} \):
\[ I(t_2) = 0.99 I_{max} = 0.99 \frac{E}{R} \]
Substituting into the equation for \( I(t) \):
\[ 0.99 \frac{E}{R} = \frac{E}{R} \left(1 - e^{-\frac{R}{L}t_2}\right) \]
Dividing both sides by \( \frac{E}{R} \):
\[ 0.99 = 1 - e^{-\frac{R}{L}t_2} \]
Rearranging gives:
\[ e^{-\frac{R}{L}t_2} = 0.01 \]

6. **Taking Natural Logarithm**:
Taking the natural logarithm on both sides:
\[ -\frac{R}{L}t_2 = \ln(0.01) \]
Thus, we find:
\[ t_2 = -\frac{L}{R} \ln(0.01) \]

7. **Finding the Ratio \( \frac{t_1}{t_2} \)**:
Now we can find the ratio:
\[ \frac{t_1}{t_2} = \frac{-\frac{L}{R} \ln(0.1)}{-\frac{L}{R} \ln(0.01)} \]
The \( -\frac{L}{R} \) cancels out:
\[ \frac{t_1}{t_2} = \frac{\ln(0.1)}{\ln(0.01)} \]
We know that \( \ln(0.1) = \ln(10^{-1}) = -\ln(10) \) and \( \ln(0.01) = \ln(10^{-2}) = -2\ln(10) \):
\[ \frac{t_1}{t_2} = \frac{-\ln(10)}{-2\ln(10)} = \frac{1}{2} \]

### Final Answer:
\[ \frac{t_1}{t_2} = \frac{1}{2} \]