A `35.0 V` battery with negligible internal resistance, a `50.0Omega` resistor, and a `1.25 mH` inductor with negligible resistance are all connected in series with an open switch. The switch is suddenly closed
(a) How long after closing the switch will the current through the inductor reach one-half of its maximum value?
(b) How long after closing the switch will the energy stored in the inductor reach one-half of its maximum value?

Explanation

To solve the problem step by step, we will address both parts (a) and (b) separately.

### Part (a): Time for Current to Reach One-Half of Its Maximum Value

1. **Identify the Maximum Current (I₀)**:
The maximum current in the circuit can be calculated using Ohm's Law:
\[ I_0 = \frac{V}{R} \]
where \( V = 35.0 \, V \) and \( R = 50.0 \, \Omega \).
\[ I_0 = \frac{35.0}{50.0} = 0.7 \, A \]

2. **Determine the Time Constant (τ)**:
The time constant for an RL circuit is given by:
\[ \tau = \frac{L}{R} \]
where \( L = 1.25 \, mH = 1.25 \times 10^{-3} \, H \) and \( R = 50.0 \, \Omega \).
\[ \tau = \frac{1.25 \times 10^{-3}}{50.0} = 2.5 \times 10^{-5} \, s \]

3. **Find Time for Current to Reach Half of Maximum Value**:
The current through the inductor at time \( t \) is given by:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
To find the time when the current reaches half of its maximum value:
\[ \frac{I_0}{2} = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
Dividing both sides by \( I_0 \):
\[ \frac{1}{2} = 1 - e^{-\frac{t}{\tau}} \]
Rearranging gives:
\[ e^{-\frac{t}{\tau}} = \frac{1}{2} \]
Taking the natural logarithm:
\[ -\frac{t}{\tau} = \ln\left(\frac{1}{2}\right) \]
Thus:
\[ t = -\tau \ln\left(\frac{1}{2}\right) = \tau \ln(2) \]

4. **Calculate Time**:
Substituting the value of \( \tau \):
\[ t = (2.5 \times 10^{-5}) \ln(2) \approx 2.5 \times 10^{-5} \times 0.693 \approx 1.73 \times 10^{-5} \, s \]

### Part (b): Time for Energy to Reach One-Half of Its Maximum Value

1. **Determine Maximum Energy (U₀)**:
The energy stored in the inductor is given by:
\[ U = \frac{1}{2} L I^2 \]
The maximum energy when the current is at its maximum \( I_0 \):
\[ U_0 = \frac{1}{2} L I_0^2 \]

2. **Find Current for Half Maximum Energy**:
For the energy to be half of its maximum:
\[ \frac{U_0}{2} = \frac{1}{2} L I^2 \]
This implies:
\[ I = \frac{I_0}{\sqrt{2}} \]

3. **Use Current Equation**:
We know:
\[ I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
Setting \( I(t) = \frac{I_0}{\sqrt{2}} \):
\[ \frac{I_0}{\sqrt{2}} = I_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
Dividing by \( I_0 \):
\[ \frac{1}{\sqrt{2}} = 1 - e^{-\frac{t}{\tau}} \]
Rearranging gives:
\[ e^{-\frac{t}{\tau}} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \]
Taking the natural logarithm:
\[ -\frac{t}{\tau} = \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]
Thus:
\[ t = -\tau \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]

4. **Calculate Time**:
Substituting the value of \( \tau \):
\[ t = (2.5 \times 10^{-5}) \ln\left(\frac{\sqrt{2} - 1}{\sqrt{2}}\right) \]
Approximating:
\[ t \approx 30.7 \times 10^{-6} \, s \]

### Final Answers:
(a) The time for the current to reach one-half of its maximum value is approximately \( 1.73 \times 10^{-5} \, s \).

(b) The time for the energy stored in the inductor to reach one-half of its maximum value is approximately \( 30.7 \times 10^{-6} \, s \).