Topic: Kirchhoff’s Law

  • Kirchhoff’s Law is named after Gustav Kirchhoff, a German physicist.
  • Kirchhoff’s Law is used to analyze complex electric circuits.
  • The two laws are:
    • Kirchhoff’s Current Law (KCL)
    • Kirchhoff’s Voltage Law (KVL)
  • These laws are based on the principles of conservation of charge and energy.

Kirchhoff’s Current Law (KCL)

  • KCL states that the algebraic sum of currents entering a junction (or a node) in an electric circuit is zero.
  • Mathematically, for a junction:
    • ∑Iin = ∑Iout
  • KCL can be used to determine unknown currents in a circuit.

Kirchhoff’s Voltage Law (KVL)

  • KVL states that the algebraic sum of voltages around any closed loop in an electric circuit is zero.
  • Mathematically, for a loop:
    • ∑Vloop = 0
  • KVL can be used to determine unknown voltages in a circuit.

Kirchhoff’s Loop Example

  • Let’s consider a simple example of a circuit consisting of a battery (EMF), resistors, and a switch.
  • We will apply KVL to analyze this circuit.
  • By applying KVL to the closed loop formed by the resistors and the battery, we can determine the current flowing through each resistor.

Use of Symmetry

  • Symmetry can be useful in simplifying circuit analysis using Kirchhoff’s laws.
  • If a circuit has symmetrical components, we can exploit the symmetry to reduce calculations.
  • By recognizing symmetrical patterns, we can analyze circuits efficiently.

Example: Symmetrical Circuit

  • Consider a circuit with two parallel resistors symmetrically connected to a common node.
  • By exploiting the symmetry, we can analyze this circuit more easily.
  • The currents in the parallel resistors are equal due to the symmetry.

Example: Symmetrical Circuit (Continued)

  • Suppose we have a circuit with two resistors R1 and R2 connected in parallel, each receiving a current I.
  • Using Ohm’s law, we can find the voltage across each resistor as V = IR.
  • The total current entering the common node is 2I.

Example: Symmetrical Circuit (Continued)

  • Applying KCL at the common node, the total current entering (2I) must be equal to the sum of currents leaving the node.
  • So, 2I = I + I, which simplifies to 2I = 2I.
  • This confirms the validity of the symmetry.

Example: Loop Analysis

  • Let’s consider another circuit with resistors and a voltage source.
  • We will apply KVL to determine the unknown currents and voltages.
  • By applying KVL to each loop, we can form a system of equations to solve for the unknowns.

Example: Loop Analysis (Continued)

  • Suppose we have a circuit with three loops and three unknown currents: I1, I2, and I3.
  • Applying KVL to each loop, we can form a system of equations:
    • Loop 1: 10V - 2I1 + I2 + I3 = 0
    • Loop 2: -3V + I1 - 2I2 + I3 = 0
    • Loop 3: 5V + I1 - I2 - 3I3 = 0
  • Solving these equations will provide the values of the unknown currents.

Kirchhoff’s Law - Kirchhoff’s Loop Example - Use of Symmetry

  • Kirchhoff’s laws can be used to analyze complex circuits.
  • We can apply Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL) to solve circuits.
  • The use of symmetry in circuits can simplify the analysis process.
  • Symmetry helps in reducing calculations and identifying equal current or voltage values.
  • Let’s explore an example to understand the application of Kirchhoff’s laws and symmetry.

Example: Circuit with Symmetrical Components

  • Consider a circuit with three resistors R1, R2, and R3.
  • R1 and R2 are connected in parallel, and R3 is connected in series with the parallel combination.
  • The circuit also includes a voltage source V.
  • Let’s determine the unknown currents and voltages using Kirchhoff’s laws and symmetry.

Example: Circuit with Symmetrical Components (Continued)

  • Applying KVL to Loop 1 formed by R1 and R2:
    • V = I1 * R1 + I2 * R2 (1)
  • Applying KVL to Loop 2 formed by R3:
    • I3 * R3 = -V + I1 * R1 + I2 * R2 (2)
  • Applying KCL at Node 1 (connecting R1, R2, and R3):
    • I1 + I2 = I3 (3)
  • These three equations help us solve for the unknown currents I1, I2, and I3.

Example: Circuit with Symmetrical Components (Continued)

  • By observing the circuit, we notice the symmetry between R1 and R2.
  • According to symmetry, the voltage across R1 must be equal to the voltage across R2.
  • This symmetry helps simplify the equations by letting us replace terms.
  • We can rewrite equation (1) as:
    • V = (I1 + I2) * R1 (4)
  • Now equation (2) becomes:
    • I3 * R3 = -V + (I1 + I2) * R1 (5)

Example: Circuit with Symmetrical Components (Continued)

  • With the help of equation (3), we can substitute (I1 + I2) with I3 in equation (4):
    • V = I3 * R1 (6)
  • Substituting this value of V in equation (5):
    • I3 * R3 = -I3 * R1 + I3 * R1 (7)
  • Simplifying equation (7):
    • 2 * I3 * R1 = I3 * R3
  • Cancelling out I3:
    • 2 * R1 = R3

Example: Circuit with Symmetrical Components (Continued)

  • We have determined that 2 * R1 = R3 based on symmetry and simplification.
  • Now we can substitute this value into the equations to find the unknown currents.
  • Equation (3) becomes:
    • I1 + I2 = I3
  • Equation (6) becomes:
    • V = I3 * R1

Example: Circuit with Symmetrical Components (Continued)

  • Combining the equations, we can solve for the unknowns:
    • I1 + I2 = 2 * R1 / R3
    • V = (2 * R1 / R3) * R1
  • Once we have these values, we can calculate other parameters such as power, potential difference, etc.
  • The use of symmetry greatly simplifies the solution process and reduces the number of variables to be solved.

Summary

  • Kirchhoff’s laws, including Kirchhoff’s Voltage Law (KVL) and Kirchhoff’s Current Law (KCL), are valuable tools for analyzing complex circuits.
  • By applying KVL and KCL, we can form systems of equations to solve for unknown currents and voltages in circuits.
  • Symmetry in circuits can help simplify the analysis process by reducing calculations and identifying equivalent current or voltage values.
  • Recognizing symmetrical patterns allows us to exploit the symmetry and solve circuits more efficiently.

Example Problems

  • Solve the following circuit using Kirchhoff’s laws and symmetry: [Circuit diagram with resistors, voltage sources, and nodes]
  • Apply KVL and KCL and identify any symmetrical components to simplify the analysis.
  • Determine all unknown currents and voltages in the circuit.

Example Problems (Continued)

  • Solve the following circuit using Kirchhoff’s laws and symmetry: [Circuit diagram with resistors, voltage sources, and nodes]
  • Apply KVL and KCL and identify any symmetrical components to simplify the analysis.
  • Determine all unknown currents and voltages in the circuit.

Slide 21

  • Example Problem 1:
    • Given circuit components:
      • R1 = 5 Ω
      • R2 = 10 Ω
      • V1 = 12 V
      • V2 = 5 V
  • Solution:
    • Apply KVL to the outer loop:
      • V1 + V2 - I1 * R1 - I2 * R2 = 0
      • 12 + 5 - I1 * 5 - I2 * 10 = 0
    • Apply KCL to the upper node:
      • I1 + I2 = I3
    • Apply KVL to the inner loop:
      • I1 * R1 + I2 * R2 = I3 * R2
    • Solve the system of equations to find I1, I2, and I3.
  • Determine all unknown currents and voltages in the circuit.

Slide 22

  • Example Problem 2:
    • Given circuit components:
      • R1 = 10 Ω
      • R2 = 15 Ω
      • R3 = 20 Ω
      • V1 = 24 V
      • V2 = 12 V
  • Solution:
    • Apply KVL to the outer loop:
      • V1 - I1 * R2 - V2 = 0
      • 24 - I1 * 15 - 12 = 0
    • Apply KCL to the right node:
      • I1 = I2 + I3
    • Apply KVL to the inner loop:
      • I2 * R3 - I3 * R1 = 0
    • Solve the system of equations to find I1, I2, and I3.
  • Determine all unknown currents and voltages in the circuit.

Slide 23

  • Example Problem 3:
    • Given circuit components:
      • R1 = 8 Ω
      • R2 = 6 Ω
      • R3 = 12 Ω
      • V1 = 10 V
      • V2 = 15 V
  • Solution:

    • Apply KVL to the outer loop:

      • V1 + R1 * I1 - V2 = 0
      • 10 + 8 * I1 - 15 = 0

    • Apply KCL to the upper node:

      • I1 + I2 = I3

    • Apply KVL to the inner loop:

      • R2 * I2 - R3 * I3 = 0

    • Solve the system of equations to find I1, I2, and I3.

  • Determine all unknown currents and voltages in the circuit.

Slide 24

  • Example Problem 4:
    • Given circuit components:
      • R1 = 5 Ω
      • R2 = 10 Ω
      • R3 = 15 Ω
      • I1 = 2 A
      • I2 = 1 A
  • Solution:
    • Apply KVL to the outer loop:
      • I1 * R1 + I3 * R3 = 0
      • 2 * 5 + I3 * 15 = 0
    • Apply KCL to the lower node:
      • I3 = I2 + I4
    • Apply KVL to the inner loop:
      • I2 * R2 - I4 * R1 = 0
    • Solve the system of equations to find I3 and I4.
  • Determine all unknown currents and voltages in the circuit.

Slide 25

  • Summary:
    • Kirchhoff’s laws (KVL and KCL) are key tools for analyzing circuits.
    • Applying KVL and KCL helps determine unknown currents and voltages in circuits.
    • Symmetry in circuits can simplify analysis by reducing the number of variables to be solved.
    • Examples of circuits with symmetrical components were discussed.
    • Solving circuits with Kirchhoff’s laws and symmetry involves forming and solving systems of equations.
    • Practice problems were provided to reinforce the concepts.