Math Problem Statement

Using Kirchhoff's laws, determine the current intensities i1, i2, and i3 in the given circuit with two voltage sources (8V and 9V) and three resistors (2Ω, 3Ω, and 4Ω).

Solution

This problem involves applying Kirchhoff's laws to determine the current intensities i1i_1, i2i_2, and i3i_3 in the given circuit.

Kirchhoff's Laws:

  1. Kirchhoff's Current Law (KCL): The sum of currents entering a junction is equal to the sum of currents leaving the junction.
  2. Kirchhoff's Voltage Law (KVL): The sum of the potential differences (voltage drops and gains) around any closed loop is zero.

Applying Ohm’s Law:

Ohm’s Law: U=IRU = IR relates the voltage UU, current II, and resistance RR.


Circuit Analysis

The circuit has two voltage sources:

  • 8V on the left, and
  • 9V on the bottom.

It has three resistors:

  • 2 ohms,
  • 3 ohms,
  • 4 ohms.

Step 1: Define Loop Currents

Let:

  • i1i_1 be the current through the 3 ohm resistor (left side),
  • i2i_2 be the current through the 2 ohm resistor (middle branch),
  • i3i_3 be the current through the 4 ohm resistor (right side).

Step 2: Kirchhoff’s Voltage Law (KVL)

We apply KVL to each loop:

Loop 1 (Left Loop):

  • Contains the 8V generator, the 2 ohm resistor, and the 3 ohm resistor.
  • The KVL equation for this loop is: 82i23i1=0(Equation 1)8 - 2i_2 - 3i_1 = 0 \quad \text{(Equation 1)}

Loop 2 (Right Loop):

  • Contains the 9V generator, the 2 ohm resistor, and the 4 ohm resistor.
  • The KVL equation for this loop is: 9+2i2+4i3=0(Equation 2)-9 + 2i_2 + 4i_3 = 0 \quad \text{(Equation 2)}

Loop 3 (Total Loop Using KCL at the junction):

At the middle node, applying Kirchhoff's Current Law: i1=i2+i3(Equation 3)i_1 = i_2 + i_3 \quad \text{(Equation 3)}

Step 3: Solve the Equations

Now we solve the system of equations:

  1. 82i23i1=08 - 2i_2 - 3i_1 = 0
  2. 9+2i2+4i3=0-9 + 2i_2 + 4i_3 = 0
  3. i1=i2+i3i_1 = i_2 + i_3

Let me calculate the values of i1i_1, i2i_2, and i3i_3.The currents in the circuit are:

  • i1=177A2.43Ai_1 = \frac{17}{7} \, \text{A} \approx 2.43 \, \text{A},
  • i2=514A0.36Ai_2 = \frac{5}{14} \, \text{A} \approx 0.36 \, \text{A},
  • i3=2914A2.07Ai_3 = \frac{29}{14} \, \text{A} \approx 2.07 \, \text{A}.

These are the intensities of the currents in the circuit as per Kirchhoff's laws and Ohm's law.

Do you have any questions or would you like further explanation?

Here are some related questions:

  1. How do Kirchhoff's laws apply to complex circuits with multiple loops?
  2. What are some common mistakes when using Kirchhoff's laws in circuit analysis?
  3. How would the current values change if one of the resistances increased?
  4. Can you explain how to identify junctions and loops in a circuit diagram?
  5. How does the direction of current affect the sign in Kirchhoff’s Voltage Law?

Tip: Always check that the sum of voltage drops equals the voltage supplied when applying KVL, and ensure all current entering and leaving a node balance for KCL!

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Math Problem Analysis

Mathematical Concepts

Circuit Analysis
Ohm's Law
Kirchhoff's Current Law (KCL)
Kirchhoff's Voltage Law (KVL)

Formulas

Ohm's Law: U = IR
Kirchhoff's Current Law: Σ I (in) = Σ I (out)
Kirchhoff's Voltage Law: Σ V = 0 in a closed loop

Theorems

Kirchhoff's Current Law
Kirchhoff's Voltage Law

Suitable Grade Level

College Level (Advanced High School Physics)