Math Problem Statement

SOLVE THIS

SOLVE THIS

Solution

This is a circuit analysis problem involving a combination of resistors, independent and dependent sources. Let's solve this step by step using nodal analysis and Ohm's law.

Step 1: Identify Nodes and Label Voltages

  • Let the node at the top of the 6Ω resistor be V₁.
  • The ground node (bottom of the circuit) is 0V.

We need to determine the voltage VxV_x, which is across the 6Ω resistor.

Step 2: Applying Nodal Analysis at V₁

From Ohm’s law, the voltage at node V₁ is:

  • Vx=V10=V1V_x = V_1 - 0 = V_1 because it's across the 6Ω resistor.

We'll now write the KCL (Kirchhoff's Current Law) equation at node V₁. The sum of currents leaving this node equals zero.

  1. Current through 6Ω resistor: I1=V1506I_1 = \frac{V_1 - 50}{6}

  2. Current through 10Ω resistor: The current source of 4 A is forcing 4 A through this path, so no need to calculate current through this branch using Ohm’s law.

  3. Current through the 15Ω resistor and dependent source: The voltage across the 15Ω resistor is V1V_1, and the current through the 15Ω resistor is: I2=V115I_2 = \frac{V_1}{15}.

The dependent current source injects a current of 2Vx=2V12V_x = 2V_1 amps into the node.

Step 3: Apply KCL at Node V₁

The sum of currents at node V1V_1 is:

V1506+4+V115=2V1\frac{V_1 - 50}{6} + 4 + \frac{V_1}{15} = 2V_1

Step 4: Solve the Equation

Let's solve for V1V_1.

Rewriting the equation: V1506+V115+4=2V1\frac{V_1 - 50}{6} + \frac{V_1}{15} + 4 = 2V_1

Multiply through by the least common denominator (LCD = 30) to clear the fractions: 5(V150)+2V1+120=60V15(V_1 - 50) + 2V_1 + 120 = 60V_1

Expand and simplify: 5V1250+2V1+120=60V15V_1 - 250 + 2V_1 + 120 = 60V_1 7V1130=60V17V_1 - 130 = 60V_1

Rearrange to solve for V1V_1: 60V17V1=13060V_1 - 7V_1 = 130 53V1=13053V_1 = 130

Finally, solve for V1V_1: V1=130532.45VV_1 = \frac{130}{53} \approx 2.45 \, \text{V}

Step 5: Determine VxV_x

Since Vx=V1V_x = V_1, we conclude: Vx=2.45VV_x = 2.45 \, \text{V}

Step 6: Current through the Dependent Source

The current through the dependent source is 2Vx=2×2.454.90A2V_x = 2 \times 2.45 \approx 4.90 \, \text{A}.

Thus, the voltage VxV_x is approximately 2.45 V, and the current through the dependent source is approximately 4.9 A.

Would you like any further details or clarifications?

Here are 5 related questions you could explore:

  1. How would the solution change if the value of the dependent current source were different?
  2. What happens to the circuit if the independent 4 A current source is removed?
  3. Can this circuit be simplified using Thevenin’s or Norton’s theorem?
  4. How would you analyze this circuit using mesh analysis instead of nodal analysis?
  5. How would you calculate the power delivered by the dependent current source?

Tip: Always double-check your equations during nodal analysis, especially when dealing with dependent sources.

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Math Problem Analysis

Mathematical Concepts

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

Formulas

I = V/R (Ohm's Law)
KCL: ΣI entering a node = ΣI leaving the node
Vx = V1 (Voltage across the 6Ω resistor)
V1 = (130/53) ≈ 2.45 V

Theorems

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

Suitable Grade Level

Undergraduate (Electrical Engineering)