Math Problem Statement

Find the values of i1, i2, and i3 in the given circuit with resistors and voltage sources.

Solution

The problem asks to find the currents i1i_1, i2i_2, and i3i_3 in the circuit, which consists of resistors and three voltage sources. Let's approach this step-by-step:

Step 1: Label the Nodes

  • Node AA connects the 5Ω, 6Ω, and 4Ω resistors.
  • Node BB connects the 8Ω resistor and 5V voltage source.
  • Node CC connects the 6Ω, 8Ω, and 16Ω resistors.
  • Node DD connects the 16Ω resistor and 2V voltage source.
  • Node EE connects the 21V source, 5Ω, and 11Ω resistors.

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

We will define the currents as follows:

  • i1i_1 flows through the 21V source and 5Ω resistor.
  • i2i_2 flows through the 5V source and 4Ω resistor.
  • i3i_3 flows through the 2V source and 16Ω resistor.

Step 3: Apply Kirchhoff's Voltage Law (KVL)

We need to write KVL equations for loops:

Loop 1: Left loop (containing the 21V source)

21V+5i1+6(i1i2)+11i1=0-21V + 5i_1 + 6(i_1 - i_2) + 11i_1 = 0

Simplifying: 22i16i2=2122i_1 - 6i_2 = 21 (Equation 1)\quad \text{(Equation 1)}

Loop 2: Right loop (containing the 5V source)

5V+4i2+6(i2i1)+8i2=05V + 4i_2 + 6(i_2 - i_1) + 8i_2 = 0

Simplifying: 6i1+18i2=5-6i_1 + 18i_2 = -5 (Equation 2)\quad \text{(Equation 2)}

Loop 3: Bottom loop (containing the 2V source)

2V+16i3+8(i3i2)=0-2V + 16i_3 + 8(i_3 - i_2) = 0

Simplifying: 24i38i2=224i_3 - 8i_2 = 2 (Equation 3)\quad \text{(Equation 3)}

Step 4: Solve the system of equations

We now have three equations:

  1. 22i16i2=2122i_1 - 6i_2 = 21
  2. 6i1+18i2=5-6i_1 + 18i_2 = -5
  3. 24i38i2=224i_3 - 8i_2 = 2

We can solve these using substitution or matrix methods.

Let me compute the solution for i1i_1, i2i_2, and i3i_3.The solutions for the currents are:

  • i1=2930A0.967Ai_1 = \frac{29}{30} \, \text{A} \approx 0.967 \, \text{A}
  • i2=245A0.0444Ai_2 = \frac{2}{45} \, \text{A} \approx 0.0444 \, \text{A}
  • i3=53540A0.0981Ai_3 = \frac{53}{540} \, \text{A} \approx 0.0981 \, \text{A}

Would you like further details or clarification on any step?

Here are 5 related questions to consider:

  1. How can Kirchhoff's Current Law (KCL) be applied at different nodes?
  2. What happens if the polarity of one of the voltage sources is reversed?
  3. How do you determine the direction of current flow in a circuit like this?
  4. Can these methods be applied to circuits with capacitors or inductors?
  5. What are the advantages of using matrix methods to solve systems of equations?

Tip: Always double-check the direction of currents when applying Kirchhoff's Laws to avoid sign errors in equations.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Electrical Circuit Analysis
Kirchhoff's Voltage Law
Kirchhoff's Current Law

Formulas

Kirchhoff's Voltage Law: ∑V = 0 around any closed loop
Ohm's Law: V = IR
Kirchhoff's Current Law: ∑I entering a junction = ∑I leaving the junction

Theorems

Kirchhoff's Voltage Law
Kirchhoff's Current Law

Suitable Grade Level

Grades 10-12 (high school physics or introductory college level electrical engineering)