Let's work through the questions provided in the circuit image.

A. Find the voltage between Point A & B.

We can calculate the voltage at points A and B using the voltage divider rule. The two resistances in series (5kΩ and 3kΩ) share the 12V total voltage, and the 4kΩ resistor is in parallel with this combination.

• First, calculate the combined resistance of the 5kΩ and 3kΩ resistors in series:

  $R_{\text{series}} = 5k\Omega + 3k\Omega = 8k\Omega$

• Now calculate the equivalent resistance of this series combination in parallel with the 4kΩ resistor:

  $\frac{1}{R_{\text{eq}}} = \frac{1}{4k\Omega} + \frac{1}{8k\Omega}$ $R_{\text{eq}} = 2.67k\Omega$

• Using Ohm's Law to find the current through the entire circuit:

  $I = \frac{12V}{R_{\text{eq}}} = \frac{12}{2.67} \approx 4.49 \, mA$

• The voltage across the 4kΩ resistor is:

  $V_{4k\Omega} = I \times 4k\Omega = 4.49 \, mA \times 4k\Omega = 17.96V$

Thus, the voltage at point A relative to point B can be determined by subtracting the potential drops along the circuit path.

Would you like me to continue solving the remaining parts B, C, D, and E? Let me know if you want more details on this part or any questions before proceeding.

Here are 5 related questions that might deepen your understanding:

1. How does the voltage divider rule apply to resistors in series?
2. What happens to the total resistance when resistors are in parallel?
3. How can Kirchhoff’s Voltage Law be used in this circuit to verify voltage at different points?
4. What is the relationship between power dissipation and voltage across each resistor?
5. How would the current change if we increased the supply voltage?

Tip: For voltage division, remember that resistors in parallel split the current inversely proportional to their resistance values.