The current in an L - R circuit in a time t = 2L / R reduces to-

Options

A. `336.5%` of maximum
B. `13.5%` of maximum
C. `0.50%` of maximum
D. `63.2%` of maximum

Answer

To solve the problem of finding the current in an L-R circuit at time \( t = \frac{2L}{R} \), we can follow these steps:

### Step 1: Understand the time constant
The time constant \( \tau \) for an L-R circuit is given by:
\[ \tau = \frac{L}{R} \]
This means that the time \( t = \frac{2L}{R} \) can be expressed in terms of \( \tau \):
\[ t = 2\tau \]

### Step 2: Use the current formula for an L-R circuit
The current \( I \) in an L-R circuit at any time \( t \) is given by the equation:
\[ I = I_0 e^{-\frac{t}{\tau}} \]
where \( I_0 \) is the maximum current, which can be expressed as:
\[ I_0 = \frac{V_0}{R} \]
where \( V_0 \) is the maximum voltage applied to the circuit.

### Step 3: Substitute \( t \) into the current formula
Now, substituting \( t = 2\tau \) into the current formula:
\[ I = I_0 e^{-\frac{2\tau}{\tau}} = I_0 e^{-2} \]

### Step 4: Calculate the ratio of current to maximum current
To find the ratio of the current \( I \) to the maximum current \( I_0 \):
\[ \frac{I}{I_0} = e^{-2} \]

### Step 5: Calculate the percentage of the maximum current
The value of \( e^{-2} \) can be approximated as:
\[ e^{-2} \approx \frac{1}{e^2} \approx \frac{1}{7.39} \approx 0.1353 \]
This means:
\[ \frac{I}{I_0} \approx 0.1353 \]
To express this as a percentage:
\[ \frac{I}{I_0} \times 100 \approx 13.53\% \]

### Final Answer
Thus, the current in the L-R circuit at time \( t = \frac{2L}{R} \) reduces to approximately **13.53%** of the maximum current \( I_0 \).

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