To solve this problem, we need to determine the optimal resistance $R$ that should be connected to the real current generator in order to maximize power transfer.

Step 1: Identify Given Data

From the problem statement:

• Voltage at a certain current: $U_1 = 100V$
• Current at this voltage: $I_1 = 25A$

Step 2: Determine the Internal Resistance

Using the linear voltage-current relationship, we assume that the generator behaves like a Thevenin equivalent circuit, where:

$U = U_{oc} - I R_{in}$

From the graph, when $I = 0$, the open-circuit voltage is:

$U_{oc} = U_1 + I_1 R_{in}$

Since at $I_1 = 25A$, $U_1 = 100V$, and at $I = I_1$, $U = 0V$:

$100V = U_{oc} - 25A \cdot R_{in}$

From the graph, the short-circuit current occurs when $U = 0$, meaning:

$I_1 = \frac{U_{oc}}{R_{in}}$

Solving for $R_{in}$:

$R_{in} = \frac{100V}{25A} = 4\Omega$

Step 3: Apply the Maximum Power Transfer Theorem

The maximum power transfer theorem states that maximum power is transferred to the load when:

$R = R_{in}$

Thus, the optimal resistance is:

$R = 4 \Omega$

Step 4: Choose the Correct Answer

Looking at the provided answer choices, the correct one is:

$R = 4\Omega$ (Option a).

Would you like a more detailed breakdown of any step? Here are some related questions for practice:

1. How do you derive the maximum power transfer theorem mathematically?
2. What happens if the connected resistance is much higher than the internal resistance?
3. How would the power transfer change if the generator's internal resistance was different?
4. Can you determine the maximum power delivered to the load in this case?
5. How would you experimentally determine $R_{in}$ in a lab setting?

Tip: When working with Thevenin circuits, always check the open-circuit voltage and short-circuit current to determine the internal resistance accurately.