The growth of current in an L-R circuit in time t=L/r is equal to about __

Options

A. `37%` of maximum
B. `63%` of maximum
C. `57%` of maximum
D. `67%` of maximum

Answer

To solve the problem of finding the growth of current in an L-R circuit at time \( t = \frac{L}{R} \), we will follow these steps:

### Step 1: Understand the L-R Circuit
An L-R circuit consists of an inductor (L) and a resistor (R) connected in series. When a voltage (V) is applied, the current does not reach its maximum value instantaneously due to the inductor's property of opposing changes in current.

### Step 2: Write the Current Expression
The expression for the current \( I(t) \) in an L-R circuit at any time \( t \) is given by:
\[ I(t) = \frac{V}{R} \left(1 - e^{-\frac{R}{L}t}\right) \]
where \( e \) is the base of the natural logarithm.

### Step 3: Substitute \( t = \frac{L}{R} \)
We need to find the current at the specific time \( t = \frac{L}{R} \). Substitute this value into the current expression:
\[ I\left(\frac{L}{R}\right) = \frac{V}{R} \left(1 - e^{-\frac{R}{L} \cdot \frac{L}{R}}\right) \]
This simplifies to:
\[ I\left(\frac{L}{R}\right) = \frac{V}{R} \left(1 - e^{-1}\right) \]

### Step 4: Calculate \( e^{-1} \)
The value of \( e^{-1} \) is approximately \( 0.3679 \). Therefore:
\[ I\left(\frac{L}{R}\right) = \frac{V}{R} \left(1 - 0.3679\right) = \frac{V}{R} \cdot 0.6321 \]

### Step 5: Determine the Growth of Current
At \( t = 0 \), the initial current \( I(0) \) is \( 0 \). The change in current from \( 0 \) to \( 0.6321 \frac{V}{R} \) is:
\[ \Delta I = I\left(\frac{L}{R}\right) - I(0) = 0.6321 \frac{V}{R} - 0 = 0.6321 \frac{V}{R} \]

### Step 6: Calculate the Percentage Growth
The percentage growth in current is given by:
\[ \text{Percentage Growth} = \left(\frac{\Delta I}{I(0) + \Delta I}\right) \times 100 = \left(\frac{0.6321 \frac{V}{R}}{0 + 0.6321 \frac{V}{R}}\right) \times 100 = 63.21\% \]

### Conclusion
Thus, the growth of current in the L-R circuit at time \( t = \frac{L}{R} \) is approximately **63%**.