A solenoid has an inductance of 50 mH and a resistance of `0.025 Omega`. If it is connected to a battery, how long will it take for the current to reach one half of its final equilibrium value?

Options

A. 1.38 s
B. 1.2 s
C. 6.32 s
D. 0.23 s

Answer

To solve the problem step by step, we will use the formula for the current in an RL circuit and the properties of logarithms.

### Step 1: Identify the given values
- Inductance (L) = 50 mH = 50 × 10^(-3) H = 0.050 H
- Resistance (R) = 0.025 Ω

### Step 2: Calculate the time constant (τ)
The time constant (τ) for an RL circuit is given by the formula:
\[ \tau = \frac{L}{R} \]
Substituting the values:
\[ \tau = \frac{0.050 \, \text{H}}{0.025 \, \Omega} = 2 \, \text{s} \]

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

### Step 4: Set up the equation for half of the final current
We want to find the time \(t\) when the current \(I(t)\) is half of its maximum value \(I_0\):
\[ I(t) = \frac{I_0}{2} \]
Substituting this into the current equation:
\[ \frac{I_0}{2} = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]

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

### Step 6: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ -\frac{t}{\tau} = \ln\left(\frac{1}{2}\right) \]
This can be rewritten as:
\[ t = -\tau \ln\left(\frac{1}{2}\right) \]

### Step 7: Substitute the value of τ
We know \(\tau = 2 \, \text{s}\):
\[ t = -2 \ln\left(\frac{1}{2}\right) \]
Using the property of logarithms \(\ln\left(\frac{1}{2}\right) = -\ln(2)\):
\[ t = 2 \ln(2) \]

### Step 8: Calculate the value of t
Using \(\ln(2) \approx 0.693\):
\[ t = 2 \times 0.693 \approx 1.386 \, \text{s} \]

### Final Answer
The time it takes for the current to reach one half of its final equilibrium value is approximately:
\[ t \approx 1.39 \, \text{s} \]