Question

Question 2 Using Superposition theorem, solve for the current through R3 in the Figure below. 1s1 = 100 mA, Vs1=28 V, R1=680 Ω, R2 = 220 Ω, R3 = 330 Ω, R4 = 533 Ω, Enter positive for down direction and negative for up direction. Enter your results in mA. R3 R2 Is1 R1 R4 4.35

Answer 1

Superposition theorem // consider one source at a time

Superposition theorem//consider one source at a time Considering Vs1 i.e. Is1 = 0, open circuit Modified figure is (a) Applying kcl V - 0/0.68 + V - (-28)/0.22 + V - 0/0.863 = 0 V = -17.73V I1 = V - 0/0.863 I1 = -20.55mA Now considering Is1 i.e. Vs1 = 0 (short circuit) Modified figure is (b) V - 0/0.68 + V - 0/0.22 + V - 0/0.863 = 100 V = 13.98 v I2 = V - 0/0.863 I2 = 16.15 mA Total current through R3 = I1 + I2 = (-20.55) + (16.15) = -4.39mA

Considering Vs1 i.e. Is1=0 , open circuit

Modified figure is (a)

Applying kcl

+ + = 0

V= -17.73V

I1=

I1= -20.55mA  

Now considering Is1 i.e. Vs1=0 (short circuit)

Modified figure is (b)

+ $\frac{V-0}{0.22}$ + = 100

V= 13.98V

I2=

I2= 16.15 mA

Total current through R3 = I1+I2

= (-20.55)+(16.15)

= -4.39mA