tailieunhanh - Lecture Electric circuits analysis - Lecture 19: Problems solving related to thevenin's theorem - Norton’s Theorem

Like Thevenin's theorem, Norton's theorem provides a method of reducing a more complex circuit to a simpler equivalent form. The basic difference is that Norton's theorem results in an equivalent current source in parallel with an equivalent resistance. Lecture provides knowledge of problems solving related to thevenin's theorem - Norton’s Theorem. | Problems Solving related to Thevenin's theorem Norton’s Theorem Lecture 19 Problems Solving-Thevenin's theorem Using Thevenin's theorem, find the voltage across R4. (Solved on 3 slides) For Vs: RTH= kΩ, VTH= V, V4= V, For IS: RTH= kΩ , I3= mA, VTH= V, V4= V, V4(total)= V For Vs: RTH= kΩ, VTH= V, V4= V, For IS: RTH= kΩ , I3= mA, VTH= V, V4= V, V4(total)= V For Vs: RTH= kΩ, VTH= V, V4= V, For IS: RTH= kΩ , I3= mA, VTH= V, V4= V, V4(total)= V Determine the current into point A when R 8 is kΩ, 5 k Ω, and 10 k Ω. (Solved on 3 slides) RTH= kΩ, RT= kΩ, IT= mA, I4= mA, V4= V, Vx= V, VTH=VA=VX = V, For R8 = 1 kΩ: IA= mA, For R8 = 15kΩ: IA= mA, For R8 = 1 kΩ: IA= mA RTH= kΩ, RT= kΩ, IT= mA, I4= mA, V4= V, Vx= V, VTH=VA=VX = V, For R8 = 1 kΩ: IA= mA, For R8 = 15kΩ: IA= mA, For R8 = 1 kΩ: IA= mA RTH= kΩ, RT= kΩ, IT= mA, I4= mA, V4= V, Vx= V, VTH=VA=VX = V, For R8 = 1 kΩ: IA= mA, For R8 = 15kΩ: IA= mA, For R8 = 1 kΩ: IA= mA Determine the Thevenin equivalent looking from terminals A and B for the given circuit. (Solved on 2 slides) VR3 = 3 V, V4=7V, R4=35 kΩ, VA= V, VB=7V, VTH= V, RTH= kΩ VR3 = 3 V, R4=35 kΩ, VA= V, VB=7V, VTH= V, RTH= kΩ Find the Thevenin equivalent for the circuit external to the amplifier in the following figure. (Solved on 2 slides) Looking back from the amplifier input RTH = Ω, For 1 V : VA= 886 mV, For 5V: VA= 200 mV, VTH = 886 mV + 200 mV = V RTH = Ω, For 1 V : VA= 886 mV, For 5V: VA= 200 mV, VTH = 886 mV + 200 mV = V NORTON'S THEOREM Like Thevenin's theorem, Norton's theorem provides a method of reducing a more complex circuit to a simpler equivalent form. The basic difference is that Norton's theorem results in an equivalent current source in parallel with an . | Problems Solving related to Thevenin's theorem Norton’s Theorem Lecture 19 Problems Solving-Thevenin's theorem Using Thevenin's theorem, find the voltage across R4. (Solved on 3 slides) For Vs: RTH= kΩ, VTH= V, V4= V, For IS: RTH= kΩ , I3= mA, VTH= V, V4= V, V4(total)= V For Vs: RTH= kΩ, VTH= V, V4= V, For IS: RTH= kΩ , I3= mA, VTH= V, V4= V, V4(total)= V For Vs: RTH= kΩ, VTH= V, V4= V, For IS: RTH= kΩ , I3= mA, VTH= V, V4= V, V4(total)= V Determine the current into point A when R 8 is kΩ, 5 k Ω, and 10 k Ω. (Solved on 3 slides) RTH= kΩ, RT= kΩ, IT= mA, I4= mA, V4= V, Vx= V, VTH=VA=VX = V, For R8 = 1 kΩ: IA= mA, For R8 = 15kΩ: IA= mA, For R8 = 1 kΩ: IA= mA RTH= kΩ, RT= kΩ, IT= mA, I4= mA, V4= V, Vx= V, VTH=VA=VX = V, For R8 = 1 kΩ: IA= mA, For R8 = 15kΩ: IA= mA, For R8 = 1 kΩ: IA= mA RTH= kΩ, .