7.09 Superposition theorem example.

Superposition theorem.

Use the superposition theorem, to find all the currents and the voltages in the circuit shown below.

superposition theorem

Circuit 1.
Current flows from E1 up through R1 and down the other branches.

superposition theorem

superposition theorem

Circuit 2.
Current flows from E2 up through R3 and down the other branches.

superposition theorem

superposition theorem

R_T = R₁ + (R₂ x R₃)/(R₂ + R₃) .
R_T = 7.75 Ohms.
I₁ = E/R_T .
I₁ = 1.29 A .
Using the current divider rule I₂ = I₁ x R₃/(R₂ + R₃) .
I₂ = 0.806 A .
I₃ = I₁ - I₂ .
I₃ = 0.484 A .

R_T = R₃ + (R₁ x R₂)/(R₁ + R₂) .
R_T = 12.4 Ohms.
I₃ = E/R_T .
I₃ = 1.21 A .
Using the current divider rule I₁ = I₃ x R₁/(R₁ + R₂) .
I₂ = 0.484 A .
I₁ = I₃ - I₂ .
I₁ = 0.726 A .

Final circuit values.

superposition theorem

superposition theorem

current in branch 1 = 1.29 A (up) - 0.726 A (down) = 0.564A (up).
current in branch 2 = 0.806 A (down) + 0.484 A (down) = 1.29A (down).
current in branch 3 = 1.21 A (up) - 0.484 A (down) = 0.726 A (up).

Checking final current values using Kirchhoff's current law applied to the junction in circuit.
I₁ + I₃ = I₂ .
0.564 + 0.726 = 1.29 A .

Calculating the voltage drops.
V₁ = I₁R₁ = 2.256V .
V₂ = I₂R₁ = 7.74V .
V₃ = I₃R₁ = 7.26V .

Checking final values using Kirchhoff's voltage law: (the voltage across each branch should be the same).
E₁ - V₁ = 7.74V .
V₂ = 7.74V .
E₂ - V₃ = 7.74V .