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.