Lecture Electric circuits analysis - Lecture 17: Problems solving-superposition theorem. In this chapter, the following content will be discussed: analysis of series-parallel resistive circuits, voltage dividers with resistive loads, ladder networks, | Problems Solving-Superposition Theorem RT(SI)=, IT(S1)=, I3(S1)= RT(S2)=, IT(S2)=, I3(S2)=. I3(tot)=910µA Find the total Current through R3 Lecture 17 RT(SI)=, IT(S1)=, I3(S1)= RT(S2)=, IT(S2)=, I3(S2)=. I3(tot)=910µA Using the superposition method, calculate the current through R 5. (Solved on 2 slides) RT(2V) = kΩ, IT= mA, I3= 577 μA, I5= 180 μA , RT(3V) = kΩ, IT= mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA RT(2V) = kΩ, IT= mA, I3= 577 μA, I5= 180 μA , RT(3V) = kΩ, IT= mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA Use the superposition theorem to find the current in and the voltage across the R2 branch of the circuit. (Solved on 2 slides) RT(2V) = kΩ, IT= mA,I2= 443 μA, RT(3V) = kΩ, IT= mA, I3= 865 μA, I2= 270 μA, I2 = 443 μA + 270 μA = 713 μA RT(2V) = kΩ, IT= mA,I2= 443 μA, RT(3V) = kΩ, IT= mA, I3= 865 μA, I2= 270 μA, I2 = 443 μA + 270 μA = 713 μA Using the superposition theorem, solve for the current through R3. (Solved on 2 slides) For Is: RT=, I3= mA For Vs:RT = Ω, IT= mA, I3= mA, I3(total) = mA For Is: RT=, I3= mA For Vs:RT = Ω, IT= mA, I3= mA, I3(total) = mA Using the superposition theorem, find the load current. IL = 361 mA, I(tot)=362mA Using the superposition theorem, find the load current. (Solved on 2 slides) For 40V: IL = 0 A, For .5A: IL=0A, For 60V: VL=, IL= mA, IL(tot) = mA For 40V: IL = 0 A, For .5A: IL=0A, For 60V: VL=, IL= mA, IL(tot) = mA Determine the voltage from point A to point B in the following Figure. (Solved on 3 slides) For 75 V: R2345= kΩ, VA=13V, VB= For 50V: R1245= 25 kΩ, VA=− V, VB=− For 100V: R123=, RT=, IT=850μA, VA= V, VB=− V ,VA= V, VB =− V, VAB = V For 75 V: R2345= kΩ, VA=13V, VB= For 50V: R1245= 25 kΩ, VA=− V, VB=− For 100V: R123=, RT=, IT=850μA, VA= V, VB=− V ,VA= V, VB =− V, VAB = V For 75 V: R2345= kΩ, VA=13V, VB= For 50V: R1245= 25 kΩ, VA=− V, VB=− For 100V: R123=, RT=, IT=850μA, VA= V, VB=− V ,VA= V, VB =− V, VAB = V In the following figure two ladder networks are shown. Determine the current provided by each of the batteries when terminals A are connected (A to A) and terminals B are connected (B to B). (Solved on 2 slides) For Vs1: RT= kΩ, IT= For Vs2: RT= kΩ, IT= mA For Vs1: RT= kΩ, IT= For Vs2: RT= kΩ, IT= mA | Problems Solving-Superposition Theorem RT(SI)=, IT(S1)=, I3(S1)= RT(S2)=, IT(S2)=, I3(S2)=. I3(tot)=910µA Find the total Current through R3 Lecture 17 RT(SI)=, IT(S1)=, I3(S1)= RT(S2)=, IT(S2)=, I3(S2)=. I3(tot)=910µA Using the superposition method, calculate the current through R 5. (Solved on 2 slides) RT(2V) = kΩ, IT= mA, I3= 577 μA, I5= 180 μA , RT(3V) = kΩ, IT= mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA RT(2V) = kΩ, IT= mA, I3= 577 μA, I5= 180 μA , RT(3V) = kΩ, IT= mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA Use the superposition theorem to find the current in and the voltage across the R2 branch of the circuit. (Solved on 2 slides) RT(2V) = kΩ, IT= mA,I2= 443 μA, RT(3V) = kΩ, IT= mA, I3= 865 μA, I2= 270 μA, I2 = 443 μA + 270 μA = 713 μA RT(2V) = kΩ, IT= mA,I2= 443 μA, RT(3V) = kΩ, IT= mA, I3=
