Figure 36 Using the Current Law

Figure 35 Illustration of Kirchhoffs Current Law

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Basic DC Theory KIRCHHOFF’S LAWS Solution: Figure 36 Using the Current Law First, apply Kirchhoff’s voltage law to both loops. Loop ABCDEF Loop ABGHEF IR = Esource IR = Esource 2 I_total + 6 I₁ = 6 2 I_total + 3 I₂ = 6 Since Kirchhoff’a current law states I_total = I₁ + I₂, substitute (I₁ + I₂) in the place of I_total in both loop equations and simplify. Loop ABCDEF Loop ABGHEF 2 (I₁ + I₂) + 6 I₁ = 6 2 (I₁ + I₂) + 3 I₂ = 6 2 I₁ + 2 I₂ + 6 I₁ = 6 2 I₁ + 2 I₂ + 3 I₂ = 6 8 I₁ + 2 I₂ = 6 2 I₁ + 5 I₂ = 6 We now have two equations and two unknowns and must eliminate I₁ to find I₂. One way is to multiply Loop ABGHEF equation by four, and subtract Loop ABCDEF equation from the result. Multiply by 4: 4 (2 I₁ + 5 I₂ = 6) 8 I₁ + 20 I₂ + 24 Rev. 0 Page 47 ES-02

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