Let's analyze the circuit step by step to answer the questions in the image:

Circuit Overview:

You have a series-parallel AC circuit where:

• $V_{\text{in}} = 10 \, \text{V}_{\text{rms}}$, 50 Hz AC source.
• The resistors are all 1 kΩ (there are five 1 kΩ resistors in total).
• You need to find $V_{\text{out}}$ across the resistor at the far right of the circuit.

Step-by-Step Solutions:

1. Find $V_{\text{out}}$:

   To solve for $V_{\text{out}}$, you need to calculate the equivalent resistance of the entire circuit first. Let's break the circuit down:

   • The first two resistors on the top are in series: $R_{\text{series1}} = 1 \, \text{k}\Omega + 1 \, \text{k}\Omega = 2 \, \text{k}\Omega$.
   • The middle resistor on the bottom (1 kΩ) is in parallel with the two series resistors: $R_{\text{parallel}} = \frac{(R_{\text{series1}} \times 1 \, \text{k}\Omega)}{(R_{\text{series1}} + 1 \, \text{k}\Omega)} = \frac{(2 \times 1)}{(2 + 1)} = \frac{2}{3} \, \text{k}\Omega$.
   • Now, add the last two resistors (both 1 kΩ) in series with this parallel combination: $R_{\text{total}} = \frac{2}{3} \, \text{k}\Omega + 1 \, \text{k}\Omega + 1 \, \text{k}\Omega = \frac{8}{3} \, \text{k}\Omega \approx 2.67 \, \text{k}\Omega.$

   The total voltage across the circuit is 10 V. Using a voltage divider, you can calculate $V_{\text{out}}$: $V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{last}}}{R_{\text{total}}} = 10 \, \text{V} \times \frac{1 \, \text{k}\Omega}{\frac{8}{3} \, \text{k}\Omega} = 10 \times \frac{3}{8} = 3.75 \, \text{V}.$

   So, $V_{\text{out}} \approx 3.75 \, \text{V}$.

2. Write a formula for $V_{\text{out}}$:

   General formula using a voltage divider: $V_{\text{out}} = V_{\text{in}} \times \frac{R_{\text{last}}}{R_{\text{total}}},$ where $R_{\text{total}}$ is the equivalent resistance of the entire circuit.

3. Use Kirchhoff, Thevenin, Norton:

   • Kirchhoff's Voltage Law (KVL): You can apply KVL around loops to verify that the sum of voltage drops equals $V_{\text{in}}$.
   • Thevenin's Theorem: You can reduce the left part of the circuit (left of the right-most resistor) into a Thevenin equivalent with a Thevenin voltage and Thevenin resistance.
   • Norton's Theorem: Similar to Thevenin’s, but you represent the circuit with an equivalent current source and parallel resistance.

4. Write algebraic form for this:

   For Thevenin equivalent:

   • Thevenin voltage $V_{\text{th}}$: Calculate the open-circuit voltage at the point where the output resistor is connected.
   • Thevenin resistance $R_{\text{th}}$: This would be the equivalent resistance seen from the terminals of the output resistor with the source replaced by its internal resistance (AC source has an internal resistance of 0 for this problem).

5. Learn about Cramer’s rule:

   Cramer’s rule is a method for solving systems of linear equations using determinants. It can be applied if you use mesh analysis to solve this circuit by forming simultaneous equations for each loop.

Questions to Explore Further:

1. How does the frequency of the AC source affect the circuit if we consider capacitors or inductors?
2. How do you apply Norton's Theorem to find the equivalent circuit?
3. What is the advantage of using Thevenin’s Theorem over direct analysis?
4. How does Kirchhoff’s Current Law (KCL) apply to the nodes in this circuit?
5. How would the circuit change if you replaced one resistor with a capacitor?

Tip: Always remember to check your final answers for consistency by summing voltage drops (KVL) or currents (KCL) in the circuit!