WebTranscribed Image Text: If Req =85 Q in the circuit of Fig. 2.105, find R. Iter R 310 30 Ω Req 12 Ω 12 Ω 60 Ω Expert Solution Want to see the full answer? Check out a sample … WebFirst you need to pick a reference, so we place a ground at the bottom of the circuit. Then we identify the unknown node and then write our nodal equations. Next we apply a constraint equation to solve for vo. v1 + − 60 V + − 120 kΩ 120 V 120 kΩ + vo 30 kΩ − At node 1, [ ( (v1–60)–0)/30k] + [ (v1–0)/120k] + [ ( (v1–120)/120k] = 0 and
Se R e q = 50 Ω no circuito da figura 2.105 , determine R
WebCalculate the current gain i o / i s in the circuit of Fig. 3.105. Figure 3.105 Chapter 3, Solution 61. At node 1, i s = (v 1 /30) + ((v 1 – v 2 )/20) which leads to 60i s = 5v 1 – 3v 2 … Web1 sep. 2024 · This is one way of checking results.Practice Problem 2.13 For the circuit shown in Fig. 2.45, find: (a) v1 and v2, (b) the power dissipated in the 3-k⍀ and 20-k⍀ resistors, and (c) the power supplied by the current source. 1 kΩ 3 kΩ + 30 mA 5 kΩ + v1 v2 20 kΩ − − Figure 2.45 For Practice Prob. 2.13. line organisation is suitable for
Solution of fundamentals of electric circuits (1) - Academia.edu
Web16 jan. 2024 · Se procede de la siguiente manera: Se halla el circuito equivalente paralelo de las resistencias 12Ω//12Ω//12Ω: Re₁ = [ (12Ω)⁻¹ + (12Ω)⁻¹ + (12Ω)⁻¹]⁻¹ = [0,0833 + … WebDownload PDF. Chapter 2, Problem 1. The voltage across a 5-kΩ resistor is 16 V. Find the current through the resistor. Chapter 2, Solution 1 v = iR i = v/R = (16/5) mA = 3.2 mA Chapter 2, Problem 2. Find the hot … Web16 sep. 2024 · La segunda ecuación, que relaciona las corrientes, es la condición de que el voltaje en el resistor sea igual al voltaje en el resistor . (2) A partir de las ecuaciones anteriores ( (1) y (2) se encuentra la corriente . Luego sustituimos esta relación en la ecuación (2), Entonces, ahora el I_1 actual se da como. line or border at which a surface terminates