Math Problem Statement

Solve these questions with final answer

Find all node voltages and find Vx in the given circuit.

Solution

To solve this circuit problem, we need to use nodal analysis and other circuit analysis techniques to find the node voltages and the value of VxV_x.

Let's go through the solution step-by-step.

1. Define Node Voltages

Let's label the top node as V1V_1, the node where VxV_x is located as V2V_2, and the ground node at the bottom.

2. Apply KCL (Kirchhoff's Current Law) at Each Node

Node V1V_1:

At node V1V_1, we have a 10V voltage source, a 3Ω resistor to V2V_2, and a dependent current source (5 times the current ixi_x) connected to ground. Applying KCL:

V1106+V1V23+5ix=0\frac{V_1 - 10}{6} + \frac{V_1 - V_2}{3} + 5i_x = 0

Since ixi_x is defined as the current through the 1Ω resistor (from V2V_2 to ground), we can express ixi_x as:

ix=V21=V2i_x = \frac{V_2}{1} = V_2

Substituting ix=V2i_x = V_2 into the equation:

V1106+V1V23+5V2=0\frac{V_1 - 10}{6} + \frac{V_1 - V_2}{3} + 5V_2 = 0

Node V2V_2:

At node V2V_2, we have a 2Ω resistor to ground, a 1Ω resistor to ground, and a dependent voltage source (2 times ixi_x), with voltage VxV_x across it. Applying KCL:

V2V13+V22+ix=0\frac{V_2 - V_1}{3} + \frac{V_2}{2} + i_x = 0

Substitute ix=V2i_x = V_2 into the equation:

V2V13+V22+V2=0\frac{V_2 - V_1}{3} + \frac{V_2}{2} + V_2 = 0

3. Solve the System of Equations

We now have two equations:

  1. V1106+V1V23+5V2=0\frac{V_1 - 10}{6} + \frac{V_1 - V_2}{3} + 5V_2 = 0
  2. V2V13+V22+V2=0\frac{V_2 - V_1}{3} + \frac{V_2}{2} + V_2 = 0

Solving these equations simultaneously will give us the values of V1V_1 and V2V_2.

Final Answer

Node Voltages:

  • V1=calculated valueV_1 = \text{calculated value}
  • V2=calculated valueV_2 = \text{calculated value}

Finding VxV_x:

Since Vx=2ixV_x = 2 \cdot i_x and ix=V2i_x = V_2, then:

Vx=2V2V_x = 2 \cdot V_2

Let me calculate these values.The solutions for the node voltages are:

  • V1=110891.236VV_1 = \frac{110}{89} \approx 1.236 \, \text{V}
  • V2=20890.225VV_2 = \frac{20}{89} \approx 0.225 \, \text{V}

Finding VxV_x:

Since Vx=2V2V_x = 2 \cdot V_2:

Vx=22089=40890.449VV_x = 2 \cdot \frac{20}{89} = \frac{40}{89} \approx 0.449 \, \text{V}

Final Answers

  1. Node Voltages:

    • V11.236VV_1 \approx 1.236 \, \text{V}
    • V20.225VV_2 \approx 0.225 \, \text{V}
  2. Vx0.449VV_x \approx 0.449 \, \text{V}

Would you like further details on any step, or have any questions?

Here are some related questions you might find helpful:

  1. How do you apply KCL and KVL in complex circuits?
  2. What are dependent sources and how do they affect circuit analysis?
  3. How is nodal analysis different from mesh analysis?
  4. How do you handle multiple sources in circuit analysis?
  5. How can circuit simulators help in verifying these results?

Tip: When working with dependent sources, always express the dependency in terms of known voltages or currents to simplify your equations.

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Math Problem Analysis

Mathematical Concepts

Circuit Analysis
Nodal Analysis
Dependent Sources
Kirchhoff's Current Law (KCL)

Formulas

Kirchhoff's Current Law (KCL): The sum of currents entering a node equals the sum of currents leaving the node.
Ohm's Law: V = IR
Dependent Source: Voltage or current source that depends on another voltage or current in the circuit

Theorems

Kirchhoff's Current Law (KCL)
Ohm's Law

Suitable Grade Level

Undergraduate (Electrical Engineering)