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Let's go step by step to solve the questions.

### (A) Voltage between Points A & B
This is a voltage divider circuit. The voltage across points A and B will be based on the potential difference across the 4kΩ resistor, as it is in parallel with the combination of the 5kΩ and 3kΩ resistors.

**Step 1: Find equivalent resistance of 5kΩ and 3kΩ in series**:
\[
R_{5k + 3k} = 5kΩ + 3kΩ = 8kΩ
\]

**Step 2: Find equivalent resistance of 4kΩ in parallel with 8kΩ**:
Using the parallel resistance formula:
\[
\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}
\]
\[
\frac{1}{R_{eq}} = \frac{1}{4kΩ} + \frac{1}{8kΩ} = \frac{2}{8kΩ} + \frac{1}{8kΩ} = \frac{3}{8kΩ}
\]
\[
R_{eq} = \frac{8kΩ}{3} = 2.67kΩ
\]

**Step 3: Voltage across A and B**:
Now, we apply the voltage divider formula to find the voltage across the parallel branch (the 4kΩ resistor):
\[
V_{AB} = V_{total} \times \frac{R_{parallel}}{R_{total}} = 12V \times \frac{2.67kΩ}{(2.67kΩ + 4kΩ)} = 12V \times \frac{2.67}{6.67}
\]
\[
V_{AB} \approx 4.8V
\]

### (B) Equivalent Resistance between A & B
We already calculated the equivalent resistance as part of the previous step:
\[
R_{eq} = 2.67kΩ
\]

### (C) IV Curve of the Circuit
The IV curve plots current \(I\) versus voltage \(V\) for the circuit. Ohm's Law \( V = IR \) applies here, where the total resistance is \(2.67kΩ\). The current can be found by:
\[
I = \frac{V}{R}
\]
For different values of \(V\), this gives corresponding values of \(I\). The curve will be linear since \(I\) and \(V\) are directly proportional, with a slope of \( \frac{1}{2.67kΩ} \).

### (D) Find an Alternate Circuit with the Same IV Characteristics
An alternate circuit with the same IV characteristics will have the same equivalent resistance. Any series or parallel combination of resistors that result in an equivalent resistance of \(2.67kΩ\) can serve this purpose. For example:
- A single resistor of \(2.67kΩ\) could replace the entire circuit.

### (E) Explanation for Why These Two Circuits Are the Same
The reason the two circuits are equivalent is that they both produce the same total current for a given voltage, which means they have the same resistance. Since the IV curve is a graphical representation of Ohm's Law, circuits with the same resistance will have the same IV curve.

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Would you like further clarification or more details on any of the solutions?  
Here are some additional questions for practice:

1. How would the voltage across A and B change if the 5kΩ resistor were removed?
2. What would happen to the current in the circuit if the 12V source were changed to 24V?
3. How can you simplify a more complex circuit to find an equivalent resistance?
4. What is the power dissipated by the 4kΩ resistor in this circuit?
5. How would the IV curve change if we added another resistor in parallel with the 4kΩ resistor?

**Tip:** Always simplify complex circuits step by step by combining resistors in series or parallel to find the overall behavior of the circuit.

Find out the voltage between Point A & B. Find the equivalent resistance between Point A & B. Draw the IV curve of this circuit. Find an alternate circuit with the same IV characteristics from the IV curve. Explain why these two circuits are the same.

Solve a circuit problem involving a 12V source, a 4kΩ resistor in parallel with a combination of 5kΩ and 3kΩ resistors. Learn how to calculate voltage between points A and B, equivalent resistance, and draw the IV curve of the circuit. Suitable for Grades 11-12 physics students.

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