A potential difference of 6V is applied to a coil of inductance 0.5H and a resistance of 4 ohm connected in series. The time taken for the current to reach half the maximum value is (in seconds)

Options

A. `(log_e 2)/(8)`
B. `(log_e 2)/(4)`
C. `(log_e 4)/(8)`
D. `(log_e 4)/(4)`

Answer

To solve the problem, we need to find the time taken for the current in an RL circuit to reach half of its maximum value when a potential difference is applied. Here are the steps to derive the solution:

### Step 1: Identify the given values
- Potential difference (V) = 6 V
- Inductance (L) = 0.5 H
- Resistance (R) = 4 Ω

### Step 2: Calculate the maximum current (I₀)
The maximum current (I₀) in the circuit can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
Substituting the values:
\[ I_0 = \frac{6 \, \text{V}}{4 \, \Omega} = 1.5 \, \text{A} \]

### Step 3: Write the equation for current in an RL circuit
The current (I) in an RL circuit as a function of time (t) is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{R}{L}t}\right) \]

### Step 4: Set up the equation for half the maximum current
To find the time when the current reaches half of its maximum value, we set:
\[ I(t) = \frac{I_0}{2} \]
Substituting for I(t):
\[ \frac{I_0}{2} = I_0 \left(1 - e^{-\frac{R}{L}t}\right) \]

### Step 5: Simplify the equation
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} = 1 - e^{-\frac{R}{L}t} \]
Rearranging gives:
\[ e^{-\frac{R}{L}t} = 1 - \frac{1}{2} = \frac{1}{2} \]

### Step 6: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ -\frac{R}{L}t = \ln\left(\frac{1}{2}\right) \]

### Step 7: Solve for time (t)
Rearranging gives:
\[ t = -\frac{L}{R} \ln\left(\frac{1}{2}\right) \]
Substituting the values of L and R:
\[ t = -\frac{0.5 \, \text{H}}{4 \, \Omega} \ln\left(\frac{1}{2}\right) \]
\[ t = -\frac{0.5}{4} \ln\left(\frac{1}{2}\right) \]
\[ t = -0.125 \ln\left(\frac{1}{2}\right) \]

### Step 8: Simplify using properties of logarithms
Using the property \( \ln\left(\frac{1}{2}\right) = -\ln(2) \):
\[ t = 0.125 \ln(2) \]

### Step 9: Final calculation
Using the approximate value \( \ln(2) \approx 0.693 \):
\[ t \approx 0.125 \times 0.693 \approx 0.086625 \, \text{s} \]

Thus, the time taken for the current to reach half the maximum value is approximately **0.0866 seconds**.