# 7.09 Superposition theorem example.

Superposition theorem.

Use the superposition theorem, to find all the currents and the voltages in the circuit shown below. Circuit 1.
Current flows from E1 up through R1 and down the other branches. Circuit 2.
Current flows from E2 up through R3 and down the other branches. • RT = R1 + (R2 x R3)/(R2 + R3) .
• RT = 7.75 Ohms.
• I1 = E/RT .
• 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 .
• RT = R3 + (R1 x R2)/(R1 + R2) .
• RT = 12.4 Ohms.
• I3 = E/RT .
• 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. • 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 .