Math Problem Statement

Determine the current in each branch using the Nodal analysis for the given circuit.

Solution

To determine the current in each branch using nodal analysis for the given circuit, let's first outline the steps:

Step 1: Identify Nodes

Let’s assign nodes for clarity:

  • Node 1: The top point between the 3Ω, 10Ω, and 5Ω resistors.
  • Ground: The bottom of the circuit (reference node).

Step 2: Apply KCL (Kirchhoff's Current Law) at each node

At each node, the sum of the currents leaving the node is zero.

For Node 1:

Let the voltage at Node 1 be V1V_1 with respect to the reference node (ground).

Using Ohm’s law for each branch, we express the currents in terms of V1V_1:

  1. Current through the 3Ω resistor: I=V13I_{\text{3Ω}} = \frac{V_1}{3}

  2. Current through the 10Ω resistor (connected between Node 1 and the 5A current source): Since there is a 5A current source on the left, the current through the 10Ω resistor must balance with that current source: I10Ω=V110I_{\text{10Ω}} = \frac{V_1}{10}

  3. Current through the 5Ω resistor (connected between Node 1 and the 10A current source): This is the current flowing through the 5Ω resistor due to the 10A current source on the right: I=V15I_{\text{5Ω}} = \frac{V_1}{5}

Now, applying KCL at Node 1:

Current entering=Current leaving\text{Current entering} = \text{Current leaving} 5A+V110=V13+V15+10A5A + \frac{V_1}{10} = \frac{V_1}{3} + \frac{V_1}{5} + 10A

Step 3: Simplify the Equation

To simplify, let’s first multiply through by the least common denominator, which is 30:

30×5+30×V110=30×V13+30×V15+30×1030 \times 5 + 30 \times \frac{V_1}{10} = 30 \times \frac{V_1}{3} + 30 \times \frac{V_1}{5} + 30 \times 10

150+3V1=10V1+6V1+300150 + 3V_1 = 10V_1 + 6V_1 + 300

Now combine like terms:

150+3V1=16V1+300150 + 3V_1 = 16V_1 + 300

150300=16V13V1150 - 300 = 16V_1 - 3V_1

150=13V1-150 = 13V_1

Solve for V1V_1:

V1=1501311.54 voltsV_1 = \frac{-150}{13} \approx -11.54 \text{ volts}

Step 4: Calculate the Currents in Each Branch

Now that we have V1V_1, we can calculate the currents through each resistor:

  1. Current through the 3Ω resistor: I=V13=11.5433.85 AI_{\text{3Ω}} = \frac{V_1}{3} = \frac{-11.54}{3} \approx -3.85 \text{ A}

  2. Current through the 10Ω resistor: I10Ω=V110=11.54101.15 AI_{\text{10Ω}} = \frac{V_1}{10} = \frac{-11.54}{10} \approx -1.15 \text{ A}

  3. Current through the 5Ω resistor: I=V15=11.5452.31 AI_{\text{5Ω}} = \frac{V_1}{5} = \frac{-11.54}{5} \approx -2.31 \text{ A}

Final Answer:

  • Current through 3Ω: 3.85A-3.85 \, \text{A} (negative indicates direction opposite to assumed).
  • Current through 10Ω: 1.15A-1.15 \, \text{A}.
  • Current through 5Ω: 2.31A-2.31 \, \text{A}.

Let me know if you'd like a detailed explanation for any of these steps!

Follow-up Questions:

  1. How is the polarity of voltages determined in nodal analysis?
  2. What happens if one of the resistances is replaced with a capacitor?
  3. How would the current values change if the 5A current source was changed?
  4. Can you explain the impact of having more nodes in this circuit?
  5. How would mesh analysis differ in solving this circuit?

Tip:

In nodal analysis, always make sure to correctly set up the ground (reference) and ensure consistency in assuming the direction of currents.

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Math Problem Analysis

Mathematical Concepts

Circuit Analysis
Nodal Analysis
Kirchhoff's Current Law (KCL)
Ohm's Law

Formulas

Ohm's Law: V = IR
Kirchhoff's Current Law: Sum of currents at a node = 0

Theorems

Kirchhoff's Current Law (KCL)

Suitable Grade Level

Grades 11-12