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The image you've provided is related to an RL circuit problem, which consists of a resistor (R) and an inductor (L) in series with a source of electromotive force (E). The goal is to model the circuit and find the general solution for the current \( I(t) \), given that \( R \) and \( L \) are constants.

### Key Data from the Problem:
- \( L = 0.25 \, H \) (Henry, inductance),
- \( R = 10 \, \Omega \) (Ohms, resistance),
- \( E = 48 \, V \) (Volts, voltage source).

### The RL Circuit Differential Equation:
The governing equation for an RL circuit is given by:

\[
L \frac{dI(t)}{dt} + RI(t) = E(t)
\]

Where:
- \( I(t) \) is the current as a function of time,
- \( E(t) \) is the applied voltage.

For this case, assuming a constant voltage \( E(t) = 48 \, V \), we can solve the equation.

### General Solution for the Current \( I(t) \):

1. The homogeneous equation (with \( E(t) = 0 \)) is:

\[
L \frac{dI(t)}{dt} + RI(t) = 0
\]

Solving this gives:

\[
I_h(t) = C e^{-\frac{R}{L}t}
\]

Where \( C \) is a constant of integration, determined by initial conditions.

2. The particular solution for \( E(t) = 48 \, V \) is a constant current solution:

\[
I_p(t) = \frac{E}{R} = \frac{48}{10} = 4 \, A
\]

3. The total solution is the sum of the homogeneous and particular solutions:

\[
I(t) = C e^{-\frac{R}{L}t} + 4
\]

4. Using an initial condition, such as \( I(0) = 0 \) (no initial current), we can find \( C \):

\[
I(0) = C + 4 = 0 \quad \Rightarrow \quad C = -4
\]

Thus, the solution for the current is:

\[
I(t) = 4(1 - e^{-\frac{10}{0.25} t}) = 4(1 - e^{-40t})
\]

### Interpretation of the Graph:
The graph in Fig. 67 shows the current \( I(t) \) as a function of time. As expected, the current starts at zero and asymptotically approaches 4 A, with an exponential decay term. The steep initial rise indicates the rapid buildup of current in the circuit due to the inductor.

### Would you like a more detailed breakdown of the steps or clarifications on any part of this solution? 

Here are five related questions for further exploration:
1. How would the current equation change if \( E(t) \) were time-dependent?
2. What happens to the current when \( L \) or \( R \) changes in value?
3. How do inductors behave in the steady-state (long-term) in RL circuits?
4. How can you determine the energy stored in the inductor over time?
5. What is the physical significance of the time constant \( \tau = \frac{L}{R} \)?

**Tip**: For RL circuits, the time constant \( \tau = \frac{L}{R} \) helps determine how quickly the current builds up. Smaller \( \tau \) means faster rise to steady-state current.

Model the RL-circuit in Fig. 66. Find the general solution when R, L, and E are any constants. Graph or sketch solutions when L = 0.25 H, R = 10 Ω, and E = 48 V.

This RL circuit problem involves finding the general solution for current I(t) in an RL circuit, given inductance L = 0.25 H, resistance R = 10 Ω, and voltage E = 48 V. The problem covers differential equations and exponential decay with a graph illustrating current behavior.

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