Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN: 9780133923605
Author: Robert L. Boylestad
Publisher: PEARSON
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Transcribed Image Text:Q3. For the power electronic circuit shown in figure below, sketch the output
voltage Vo, and Find the average & RMS output Vo & io and average ip.
VA
V mọ
as=160
ip
et
Ve
io
VAB = 415V
Vo
R = 1002
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- --..: : Vims t vLtj; loot A Sinwt be..ļ03.1 L35-S). Qlo:. Tf is Known . Detenmine the amplituch A to Ansarrow_forwardDetermi ne V, and Ve for the circuitin figure below. 2201 VIarrow_forwardCurrent Attempt in Progress In the figure R₁ = R₂=4.502, R3-7.202, R4 3.80, R5 1.302, and R6 R7 Rg=8.802, and the ideal batteries have emfs 1 = 26 V and 82-8.3 V. What are the (a) magnitude and (b) direction (up or down) of current is and the (c) magnitude and (d) direction of current i2? What is the energy transfer rate in (e) battery 1 and (f) battery 2? Is energy being supplied or absorbed in (g) battery 1 and (h) battery 2? (a) Number i (b) (c) Number (d) O (e) Number i (8) O (f) Number i (h) O O R₁ R₂ www R₂ R₂ Ro Units Units Units Units ww www R₂ R₁ 4 8,0 342 R₂ O Oarrow_forward
- Exercise 2: Calculate the following i. Current flowing through the circuit. (Irms) ii. Current flowing through the resistors R9, R10, R11. (Irms) iii. Calculate the peak to peak- peak voltage (Vpk-pk) and peak-peak current (I pk-pk) iv. Calculate the time period of the waveformarrow_forwardIn the circuit shown in the figure, Rs = 1.88Kohm, R1 = 1201.67Kohm, R2 = 5.72Kohm, R3 = 59.80Kohm, R4 = 15.44Kohm, R5 = 3.15Kohm, R6 = 1.05Kohm, Ry 94.19Kohm, VCC1 = 11.00V, %3D VCC2 = 10.00V, Beta1= 126.00, Beta2= 248.00, VT = 25mV, IVBE| = 0.7V and OPAMP is ideal, calculate the absolute value of the voltage gain (IVO/VS|) of the circuit relative to the source (The %3D result will be positive).arrow_forwardIn the circuit given in the figure | VBE | = 0.7V, R1 = 11.06Kohm, R2 = 48.24Kohm, RC = 1.03Kohm, RE = 3.09Kohm, VCC = -13.00V, Beta = 150.00, calculate the IC current by full analysis.arrow_forward
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