In an R-L series circuit, the source DC voltage is V. Find the current at time \(\rm t=\frac{L}{R}\) sec.

Explanation:

In an R-L series circuit, when a direct current (DC) voltage is applied, the current doesn't immediately reach its maximum value due to the presence of the inductor. Inductors resist changes in current, causing the current to increase gradually over time. The time constant of an R-L circuit is given by \( \tau = \frac{L}{R} \), where \( L \) is the inductance and \( R \) is the resistance.

Let's find the current at time \( t = \frac{L}{R} \) seconds.

Step-by-Step Solution:

1. **Initial Conditions and the Differential Equation**:

When a DC voltage \( V \) is applied to an R-L series circuit, the current \( I \) through the circuit at any time \( t \) is governed by the differential equation:

\( V = L \frac{dI}{dt} + IR \)

2. **Solution of the Differential Equation**:

This is a first-order linear differential equation. The solution to this equation, considering \( I(0) = 0 \) (since the current is zero at the moment the voltage is applied), is:

\( I(t) = \frac{V}{R}(1 - e^{-\frac{R}{L}t}) \)

3. **Current at \( t = \frac{L}{R} \)**:

To find the current at time \( t = \frac{L}{R} \), substitute \( t = \frac{L}{R} \) into the equation:

\( I\left(\frac{L}{R}\right) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} \cdot \frac{L}{R}}\right) \)

4. **Simplifying the Expression**:

The exponent simplifies to \( -1 \):

\( I\left(\frac{L}{R}\right) = \frac{V}{R} \left(1 - e^{-1}\right) \)

5. **Calculating \( e^{-1} \)**:

The value of \( e^{-1} \) (where \( e \) is the base of the natural logarithm) is approximately 0.368. Therefore:

\( I\left(\frac{L}{R}\right) = \frac{V}{R} \left(1 - 0.368\right) \)

6. **Final Current Calculation**:

Simplifying further:

\( I\left(\frac{L}{R}\right) = \frac{V}{R} \times 0.632 \)

Therefore, the current at time \( t = \frac{L}{R} \) seconds is:

\( I\left(\frac{L}{R}\right) = 0.632 \frac{V}{R} \) amperes.

Conclusion:

The correct answer is option 3: \( \text{0.632} \frac{V}{R} \text{ A} \).

Additional Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \( \text{0.865} \frac{V}{R} \text{ A} \)

This option is incorrect because it does not align with the derived solution for the current at \( t = \frac{L}{R} \). The value 0.865 is not representative of the current at this specific time in an R-L circuit.

Option 2: \( \text{0.982} \frac{V}{R} \text{ A} \)

This option is also incorrect. The value 0.982 would suggest a time much further along the exponential curve where the current is nearing its maximum steady-state value of \( \frac{V}{R} \), but not at \( t = \frac{L}{R} \).

Option 4: \( \text{0.950} \frac{V}{R} \text{ A} \)

This option is incorrect for similar reasons to option 2. The value 0.950 would occur at a time closer to when the current has almost reached its maximum value, not at \( t = \frac{L}{R} \).

Understanding the behavior of current in an R-L circuit is essential in electrical engineering. The exponential nature of the current increase due to the inductor's opposition to changes in current is a fundamental concept. The time constant \( \tau = \frac{L}{R} \) gives a measure of how quickly the current reaches its steady-state value. At \( t = \tau \), the current reaches approximately 63.2% of its maximum value, which is accurately reflected in the correct option.