A coil of inductance 10H and resistance `15Omega` is connected to a supply of 90V. Determine the value of current after 0.67s. How long will it take for the current to attain 50% of its final value ?

Answer

To solve the problem step by step, we will use the formula for the current in an RL circuit when connected to a DC supply.

### Given Data:
- Inductance (L) = 10 H
- Resistance (R) = 15 Ω
- Voltage (ε) = 90 V
- Time (t) = 0.67 s

### Step 1: Determine the final steady-state current (I_final)
The final steady-state current in an RL circuit can be calculated using the formula:
\[ I_{\text{final}} = \frac{\epsilon}{R} \]

Substituting the values:
\[ I_{\text{final}} = \frac{90 \, \text{V}}{15 \, \Omega} = 6 \, \text{A} \]

### Step 2: Calculate the current at t = 0.67 s
The formula for the current at any time t in an RL circuit is:
\[ I(t) = I_{\text{final}} \left(1 - e^{-\frac{R t}{L}}\right) \]

Substituting the values:
\[ I(0.67) = 6 \left(1 - e^{-\frac{15 \times 0.67}{10}}\right) \]

Calculating the exponent:
\[ -\frac{15 \times 0.67}{10} = -1.005 \]

Now, calculate \( e^{-1.005} \):
\[ e^{-1.005} \approx 0.3679 \]

Now substituting this back into the current equation:
\[ I(0.67) = 6 \left(1 - 0.3679\right) \]
\[ I(0.67) = 6 \times 0.6321 \]
\[ I(0.67) \approx 3.7926 \, \text{A} \]

### Step 3: Determine the time to reach 50% of the final current
To find the time when the current reaches 50% of its final value:
\[ I(t) = \frac{I_{\text{final}}}{2} \]

Setting up the equation:
\[ \frac{I_{\text{final}}}{2} = I_{\text{final}} \left(1 - e^{-\frac{R t}{L}}\right) \]

Cancelling \( I_{\text{final}} \) from both sides:
\[ \frac{1}{2} = 1 - e^{-\frac{R t}{L}} \]
\[ e^{-\frac{R t}{L}} = \frac{1}{2} \]

Taking the natural logarithm on both sides:
\[ -\frac{R t}{L} = \ln\left(\frac{1}{2}\right) \]
\[ \frac{R t}{L} = -\ln(2) \]

Substituting the values:
\[ t = -\frac{L \ln(2)}{R} \]
\[ t = -\frac{10 \times 0.693}{15} \]
\[ t \approx 0.462 \, \text{s} \]

### Final Answers:
1. The current after 0.67 seconds is approximately **3.79 A**.
2. The time taken to reach 50% of the final current is approximately **0.462 s**.