Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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- 25. Consider the circuit diagram depicted in the figure. R₂ 2.5- E1-18 V R₁ a M 0.5 Ω E₂ = 45 V 0.50 R₁₂ 1.50 (a) If the current through the top branch is 12 = 0.02 A, what is the current through the bottom, /3, in amps? [Ans: 13=13.53 A]arrow_forwardThank you in advance for your answer.arrow_forwardAssume that the ideal op amp in the circuit in the figure is operating in its linear region, Take that R₁ = 45 k and R₂ = 18 kn. Part A Calculate the power delivered to the R₂. Express your answer with the appropriate units. ANSWER: PR₂ Express your answer with the appropriate units. ANSWER: PR₂ Part C Find the ratio of the power found in Part A to that found in Part B. ANSWER: 320mV Part B Calculate the power delivered to the R₂ with the op amp removed from the circuit, that is, with the R₂ resistor connected in the series with the voltage source and the R₁ resistor. PA/PB = + Source: Loadarrow_forward
- 10) i need fast please nowarrow_forwardVa=3V and Vb=0V so I think that this would result in an inverting op-amp and has the equation -(160/20)*3=-24V but this is wrong I tried the long way as well but got the same answer. The possible source of error could be the 5k resistor but that doesn't make to much sense to me. How do I solve this?arrow_forward1. Consider the ideal op-amp circuit below. Let Us = 0.1 V, R1 = R2 = R3 = R3 = 100 kN, and R4 = 98 kN, Solve for Vo. R1 RA R5 vo R2 Us R3arrow_forward
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