Kirchoff's Laws - Comprehensive Notes

Kirchoff's Laws

• Junction (Node):
  • Refers to the intersection of three or more conductors.
• Loop:
  • Any closed conducting path.

Kirchoff's Voltage Law (KVL)

• Also known as Kirchoff's Loop Rule.
• States that the algebraic sum of voltages in each loop is zero.
• $\sum V = 0$
• $V<em>1 + V</em>2 + V<em>3 + \dots + V</em>n = 0$
• $I<em>1R</em>1 + I<em>2R</em>2 + I<em>3R</em>3 + \dots + I<em>nR</em>n = 0$
• Example:
  • Around the first loop: $+ I<em>1R</em>1 + I<em>2R</em>2 - Vs = 0$
  • The sign of each term matches the polarity encountered first.
  • Around the second loop: $-I<em>2R</em>2 + I<em>3R</em>3 = 0$

Kirchhoff’s Current Law (KCL)

• Also known as Kirchoff’s Junction (Circuit) Rule.
• States that the algebraic sum of currents entering a node is zero.
• $\sum I = 0$
• For a battery with currents $I<em>1$, $I</em>2$, and $I<em>3$: $I</em>1 - (I<em>2 + I</em>3) = 0 \implies I<em>1 - I</em>2 - I_3 = 0$
• Example:
  • At node B: $I<em>1 - I</em>2 - I_3 = 0$
  • Add currents in, subtract currents out.

Example Problem

• Given:
  • $R_1 = 7 \Omega$
  • $R_2 = 10 \Omega$
  • $R_3 = 6 \Omega$
  • $R_4 = 4 \Omega$
  • $V = 24V$
• KVL Equations (2 loops):
  • Loop 1: $I<em>1R</em>1 + I<em>2R</em>2 - V = 0 \implies 7I<em>1 + 10I</em>2 - 24 = 0$
  • Loop 2: $I<em>3R</em>3 + I<em>4R</em>4 - I<em>2R</em>2 = 0 \implies 6I<em>3 + 4I</em>4 - 10I_2 = 0$
• KCL Equation:
  • $I<em>1 - I</em>2 - I_3 = 0$
• Solving the system:
  • Using $I<em>3 = I</em>4$ in loop 2 equation: $6I<em>3 + 4I</em>3 - 10I<em>2 = 0 \implies 10I</em>3 = 10I<em>2 \implies I</em>3 = I_2$
  • Using $I<em>1 = 2I</em>2$
  • Substituting into loop 1 equation: $7(2I<em>2) + 10I</em>2 - 24 = 0 \implies 14I<em>2 + 10I</em>2 - 24 = 0 \implies 24I<em>2 = 24 \implies I</em>2 = 1A$
  • Therefore:
    • $I_1 = 2(1) = 2A$
    • $I_3 = 1A$
    • $I_4 = 1A$
• Calculating Voltages (using $V = IR$):
  • $V<em>1 = I</em>1R_1 = 2A * 7\Omega = 14V$
  • $V<em>2 = I</em>2R_2 = 1A * 10\Omega = 10V$
  • $V<em>3 = I</em>3R_3 = 1A * 6\Omega = 6V$
  • $V<em>4 = I</em>4R_4 = 1A * 4\Omega = 4V$

Example 2

• Given:

  • Resistors: 2, 6, 7, 1, 10, 8
  • Voltage Source: 20V

• KVL Equations:

  • Loop 1: $I<em>1R</em>1 + I<em>2R</em>2 - 20 = 0$
  • Loop 2: $I<em>3R</em>3 + I<em>4R</em>4 + I<em>5R</em>5 - I<em>2R</em>2 = 0$

• KCL Equation:

  • $I<em>1 - I</em>2 - I_3 = 0$

• Simplifications:

  • $I<em>3 = I</em>4 = I_5$
  • $I<em>1 = I</em>6$

• Solving:

  • Using $I<em>3 = I</em>4 = I<em>5$: $7I</em>3 + 1I<em>3 + 10I</em>3 - 6I<em>2 = 0 \implies 18I</em>3 -6I<em>2=0\implies -6I</em>2 = -18I<em>3\implies I</em>2 = 3I_3$

  • Substituting into KCL: $I<em>1 = 4I</em>3$

  • Substituting into Loop 1:$V<em>1 + V</em>2 + V<em>6 - 20 = 0 \implies 2I</em>1 + 6I<em>2 + 8I</em>6 - 20 = 0 \implies 2(4I<em>3) + 6(3I</em>3) + 8(4I<em>3) - 20 \implies 8I</em>3 + 18I<em>3 + 32I</em>3 - 20 = 0 \implies 58I<em>3 = 20 \implies I</em>3 = 0.34 A$

• Therefore:

  • $I_1 = 4 * 0.34 = 1.36A$

  • $I_2 = 3 * 0.34 = 1.02A$

  • $I_4 = 0.34A$

  • $I_5 = 0.34A$

  • $I_6 = 1.36A$

• Calculating Voltages (using V = IR):

  • $V_1 = 2.72 V$

  • $V_2 = 6.12 V$

  • $V_3 = 2.38 V$

  • $V_4 = 0.34 V$

  • $V_5 = 3.4 V$

  • $V_6 = 10.88 V$