To solve the problem, we need to find \( \frac{di}{dt} \) and \( L \frac{di}{dt} \) at \( t = 50 \, \text{ms} \) for an LR circuit with given values of inductance \( L = 1.0 \, \text{H} \) and resistance \( R = 20 \, \Omega \), connected across an EMF of \( V = 2.0 \, \text{V} \).
### Step-by-Step Solution:
**Step 1: Calculate the time constant \( \tau \)**
The time constant \( \tau \) for an LR circuit is given by the formula:
\[
\tau = \frac{L}{R}
\]
Substituting the given values:
\[
\tau = \frac{1.0 \, \text{H}}{20 \, \Omega} = 0.05 \, \text{s}
\]
**Step 2: Calculate the maximum current \( I_0 \)**
The maximum current \( I_0 \) can be calculated using Ohm's law:
\[
I_0 = \frac{V}{R}
\]
Substituting the given values:
\[
I_0 = \frac{2.0 \, \text{V}}{20 \, \Omega} = 0.1 \, \text{A}
\]
**Step 3: Write the expression for current \( I(t) \)**
The current \( I(t) \) in the circuit as a function of time is given by:
\[
I(t) = I_0 \left(1 - e^{-\frac{t}{\tau}}\right)
\]
Substituting \( I_0 \) and \( \tau \):
\[
I(t) = 0.1 \left(1 - e^{-\frac{t}{0.05}}\right)
\]
**Step 4: Differentiate \( I(t) \) to find \( \frac{di}{dt} \)**
To find \( \frac{di}{dt} \), we differentiate \( I(t) \):
\[
\frac{di}{dt} = \frac{d}{dt} \left[0.1 \left(1 - e^{-\frac{t}{0.05}}\right)\right]
\]
Using the chain rule:
\[
\frac{di}{dt} = 0.1 \cdot \left(0 - \left(-\frac{1}{0.05} e^{-\frac{t}{0.05}}\right)\right) = \frac{0.1}{0.05} e^{-\frac{t}{0.05}} = 2 e^{-\frac{t}{0.05}}
\]
**Step 5: Substitute \( t = 50 \, \text{ms} \) into \( \frac{di}{dt} \)**
Convert \( t \) to seconds:
\[
t = 50 \, \text{ms} = 0.05 \, \text{s}
\]
Now substitute \( t \) into the expression for \( \frac{di}{dt} \):
\[
\frac{di}{dt} = 2 e^{-\frac{0.05}{0.05}} = 2 e^{-1}
\]
**Step 6: Calculate \( L \frac{di}{dt} \)**
Now, we can calculate \( L \frac{di}{dt} \):
\[
L \frac{di}{dt} = 1.0 \cdot (2 e^{-1}) = 2 e^{-1}
\]
### Final Answers:
- \( \frac{di}{dt} = 2 e^{-1} \, \text{A/s} \)
- \( L \frac{di}{dt} = 2 e^{-1} \, \text{V} \)
