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- Norton's Theorem. VT=14V,R1=6ohms,R2=8ohms,R3=10ohms. Nortonize this circuit for R3 as load resistance by finding IN & RN then find IL & VL. Check results by finding I3 & V3Question: In the circuit to the right, R1 = R2,W/L = 3, Vtn = 0.6 V, and Kn = 200 A/V2. Determine the values of R1 and R2 such that VD = 1 V and Vs = -1 V.1. Circuit Topology (Show solution) a. Simplify the circuit below such that there is only one voltage source and one equivalent resistor Rs in the circuit. Do not use source transformation. What is Rs? b. Simplify the circuit below such that there is only one current source I1 and one equivalent resistor Rp in the circuit. Do not use source transformation. What is Rp?
- Find the Norton equivalent resistance RN that drives R4 when E=6 V, R1=12 , R2=15, R3=33 and R4=7.Given: VR1=12V VR2=12V PR3=12mV Find: RT, R1, R2, R3 IT, IR1, IR2, IR3 VT, VR3 PR1, PR2Arm resistances of Wheatstone bridge are,R₁ = 200 £2R₂ = 400 2R₁ = 500 SR₁ = 600Input is, E5V. Find the bridge output voltage.
- (a) Draw its Norton equivalent circuit. (b) Find the load current if RL is 4 Ω. (c) Calculate the power transferred to RL. (d) Determine the load power efficiency. (e) Is the power delivered to the load maximum? Briefly explain your answer. (f) If RL is replaced with a 10 Ω potentiometer, sketch the load power graph.Where: R1= 6 Ω, R2= 10 Ω, R3= 8 Ω, R4= 6 Ω, R5= 5 Ω, R6= 7 Ω E= 22 V, I = 4 A Using Nodal analysis, Node b equation can be written as: Node b: (-1/10) Va + c1 Vb + c2 Vc = c3 if Va, Vb, and Vc are in Volts, calculate the coefficient c3.Can you combine all of the tables below onto a graph that Vgs plotted on Vds vs Id that's similar to the graph picture attached? Id1 is for table 1 where V2 = 2.5 V. Id2 is for table 2 where V2 = 3 V. Id3 is for table 3 where V2 = 3.5 V. Id4 is for table 4 where V2 = 4 V. V2 = 2.5 V V1 (V) Vds (mV) Vgs (V) Id (mA) 0 -0.000993 2.475 -0.000022 1 78 2.392 0.838 2 946 2.38 0.958 3 1946 2.38 0.958 V2 = 3 V V1 (V) Vds (mV) Vgs (V) Id (mA) 0 -0.000418 2.97 -0.000027 1 16 2.882 0.894 2 38 2.794 1.784 3 69 2.706 2.665 4 121 2.621 3.526 5 281 2.545 4.29 6 1269 2.544 4.3 7 2269 2.544 4.301 8 3269 2.544 4.3 9 4269 2.544 4.301…
- Circuit Analysis , answer the questions below for the given circuit in the picture. Thanks. V1=7V, R1=6 Ω, R2=8 Ω, R3=5 Ω, R4=5 Ω and I1=11 A i1=? PV1 =? PI1 =? P5i1 =?Subject: CircuitsGivenRt = 37.9529 ohm..Rcd = 3.6 ohm.Rab=2.3529ohmIt=1.58APt=94.85WV2=3.16VV4=6.32VVab=3.7175vVcd=5.688VI10=0.372AI5=0.7435AI8=0.465AI9=0.632AI6=0.348AP2=4.9928WP4=9.9856WPab=5.88WP14=34.95WPcd=8.99WP12=29.96WIn the circuit below, the voltage sourse Vin = 12(V), and node D is grounded. Given R1 = 12 (Ω), R2 = 16 (Ω), R3 = 16 (Ω), and R4 = 16 (Ω). Find the voltage dropped cross R2, VA-VB, in volts.