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Lecture Electric circuits analysis - Lecture 17: Problems solving-superposition theorem

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)=1.69KΩ, IT(S1)=11.8mA, I3(S1)= 3.69mA RT(S2)=1.69KΩ, IT(S2)=8.88mA, I3(S2)=2.78mA. I3(tot)=910µA Q.1 Find the total Current through R3 Lecture 17 RT(SI)=1.69KΩ, IT(S1)=11.8mA, I3(S1)= 3.69mA RT(S2)=1.69KΩ, IT(S2)=8.88mA, I3(S2)=2.78mA. I3(tot)=910µA Q.2 Using the superposition method, calculate the current through R 5. (Solved on 2 slides) RT(2V) = 1.955 kΩ, IT=1.02 mA, I3= 577 μA, I5= 180 μA , RT(3V) = 1.955 kΩ, IT= 1.53 mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA RT(2V) = 1.955 kΩ, IT=1.02 mA, I3= 577 μA, I5= 180 μA , RT(3V) = 1.955 kΩ, IT= 1.53 mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA Q.3 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) = 1.955 kΩ, IT=1.02 mA,I2= 443 μA, RT(3V) = 1.955 kΩ, IT= 1.53 mA, I3= 865 μA, I2= 270 μA, I2 = 443 μA + 270 μA = 713 μA RT(2V) = 1.955 kΩ, IT=1.02 mA,I2= 443 μA, RT(3V) = 1.955 kΩ, IT= 1.53 mA, I3= 865 μA, I2= 270 μA, I2 = 443 μA + 270 μA = 713 μA Q.4 Using the superposition theorem, solve for the current through R3. (Solved on 2 slides) For Is: RT=137.6Ω, I3=17.2 mA For Vs:RT = 587.6 Ω, IT=34.0 mA, I3= 15.6 mA, I3(total) =1.6 mA For Is: RT=137.6Ω, I3=17.2 mA For Vs:RT = 587.6 Ω, IT=34.0 mA, I3= 15.6 mA, I3(total) =1.6 mA Q.5 Using the superposition theorem, find the load current. IL = 361 mA, I(tot)=362mA Q.6 Using the superposition theorem, find the load current. (Solved on 2 slides) For 40V: IL = 0 A, For .5A: IL=0A, For 60V: VL=43.7V, IL=29.1 mA, IL(tot) = 29.1 mA For 40V: IL = 0 A, For .5A: IL=0A, For 60V: VL=43.7V, IL=29.1 mA, IL(tot) = 29.1 mA Q.7 Determine the voltage from point A to point B in the following Figure. (Solved on 3 slides) For 75 V: R2345= 17.2 kΩ, VA=13V, VB=11.7V For 50V: R1245= 25 kΩ, VA=−21.6 V, VB=−19.5V For 100V: R123=16.6kΩ, RT=117.6kΩ, IT=850μA, VA=14.1 V, VB=−77.4 V ,VA= 5.5 V, VB =−85.2 V, VAB = 90.7 V For 75 V: R2345= 17.2 kΩ, VA=13V, VB=11.7V For 50V: R1245= 25 kΩ, VA=−21.6 V, VB=−19.5V For 100V: R123=16.6kΩ, RT=117.6kΩ, IT=850μA, VA=14.1 V, VB=−77.4 V ,VA= 5.5 V, VB =−85.2 V, VAB = 90.7 V For 75 V: R2345= 17.2 kΩ, VA=13V, VB=11.7V For 50V: R1245= 25 kΩ, VA=−21.6 V, VB=−19.5V For 100V: R123=16.6kΩ, RT=117.6kΩ, IT=850μA, VA=14.1 V, VB=−77.4 V ,VA= 5.5 V, VB =−85.2 V, VAB = 90.7 V Q.8 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=14.0 kΩ, IT=2.28mA For Vs2: RT= 11.1 kΩ, IT= 1.35 mA For Vs1: RT=14.0 kΩ, IT=2.28mA For Vs2: RT= 11.1 kΩ, IT= 1.35 mA | Problems Solving-Superposition Theorem RT(SI)=1.69KΩ, IT(S1)=11.8mA, I3(S1)= 3.69mA RT(S2)=1.69KΩ, IT(S2)=8.88mA, I3(S2)=2.78mA. I3(tot)=910µA Q.1 Find the total Current through R3 Lecture 17 RT(SI)=1.69KΩ, IT(S1)=11.8mA, I3(S1)= 3.69mA RT(S2)=1.69KΩ, IT(S2)=8.88mA, I3(S2)=2.78mA. I3(tot)=910µA Q.2 Using the superposition method, calculate the current through R 5. (Solved on 2 slides) RT(2V) = 1.955 kΩ, IT=1.02 mA, I3= 577 μA, I5= 180 μA , RT(3V) = 1.955 kΩ, IT= 1.53 mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA RT(2V) = 1.955 kΩ, IT=1.02 mA, I3= 577 μA, I5= 180 μA , RT(3V) = 1.955 kΩ, IT= 1.53 mA, I5= 655 μA, I5(total) = 180 μA + 665 μA = 845 μA Q.3 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) = 1.955 kΩ, IT=1.02 mA,I2= 443 μA, RT(3V) = 1.955 kΩ, IT= 1.53 mA, I3= 865 μA, I2= 270 μA, I2 = 443 μA + 270 μA = 713 μA RT(2V) = 1.955 kΩ, IT=1.02 mA,I2= 443 μA, RT(3V) = 1.955 kΩ, IT= 1.53 mA, I3=