9. In the network shown in Figure 9 (a) and 9 (b), find the value of Rµ necessary to obtain maximum power in RL and the maximum power dissipated in RL. [Ans: a) 6 2, 0.667 W; b) 150 N, (25/24)W] 300 2 25 2 12 2 2Ω 200 6 Ω, RL 100 V RL 12 V 50 Ω Figure 9 (a) Figure 9 (b)

9. In the network shown in Figure 9 (a) and 9 (b), find the value of Rµ necessary to obtain maximum power in RL and the maximum power dissipated in RL. [Ans: a) 6 2, 0.667 W; b) 150 N, (25/24)W] 300 2 25 2 12 2 2Ω 200 6 Ω, RL 100 V RL 12 V 50 Ω Figure 9 (a) Figure 9 (b)

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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R1 = 1 kohms R2 = 1.4 kohms R3 = 6.8 kohms E = 8VD1: Si, rB = 2 ohms, rR = 215 kohms D2: Si, rB = 7 ohms, rR = 590 kohmsApplying first approximations, determine: a) Current through D1 b) Voltage across D_2 c) Voltage across R_3
For RL=180 ohms RL=470 ohms, find the VL, IL, IZ, IR voltages and currents and PZ power values separately. Briefly describe the operation of the circuit.
In the circuit given in the figure, Rk = 1.87Kohm, R1 = 4217.50Kohm, R2 = 5782.50Kohm, R3 = 1.55Kohm, R4 = 0.52Kohm, R5 = 24.75ohm, Ry = 98.79Kohm, VCC = 10.00V, VTN = 2.85V, Kn = Since it is 2.48(mA/V^2), C1= 7.22uF, C2=11.03uF, C3=14.19uF, calculate the lower cutoff frequency of the circuit in Hz.
Write an expression for the output voltage vo(t) ofthe circuit shown if R1 = 1 k , R2 =2 k , R3 =47 k , v2(t)=(0.01 sin 3770t) V, andv1(t)=(0.04 sin 10000t)V. Write an expression forthe voltage appearing at the summing junction (v−).
In the DC Circuits lab , "Ohm's Law", the resistance of the four load were measured and also their accepted values were calculated: For Load A: Resisitance=4.4 kΩ (accepted value) caluculated using formula =R= R1+ R2 + R3 Resisitance=4.42kΩ (experimental value) For Load B: Resisitance=0.47 kΩ (accepted value) caluculated using formula =R= 1/(1/R1+ 1/R2 + 1/R3) Resisitance=0.47kΩ (experimental value) For Load C: Resisitance=1.95 kΩ (accepted value) caluculated using formula =R= R1+(( R2*R3)/R2 +R3) Resisitance=1.95kΩ (experimental value) For Load D: Resisitance=1.08 kΩ (accepted value) caluculated using formula =R= ((R1+ R2)* R3)/R1+R2+R3 Resisitance=1.08kΩ (experimental value) Percent Discrepancy for each load were 0 What's the conclusion then?
Given a series circuit consisting of a DC voltage source, resistor, capacitor, inductor, and switch which closes at t=0. All elements are initially uncharged. If V=4 volts, R=64 Ohms, C=1/520 Farads, and L=8 Henries, Determine the circuit current (A,B,H,D): i(t)=Aexp(-Bt)sin(Ht+D) amps.
A 3-ph, 50 Hz, 150 km length O.H.T.L. supplies a load with P 44 MW load at 0.85 lag P.f. and 69 kV This T.L has a resistance of 0.232 Ω/km and an inductive reactance of 0.783 Ω/km and thecapacitance to neutral is 0.0446 μF/km. Using nominal π circuit, Find:1- Constants A, B, C, and D. 2- Sending end voltage. 3- Sending end current4- Voltage regulation of this T.L. 5- Transmission line efficiency.6- Draw with a scale the phasor diagram of this model at this operating case.7- Estimate approximately the loadability of this transmission line at this operating voltage. Justifyyour estimation and suggest a way to increase this loadability. Verify your answer.
In the circuit given in the figure, Rk = 1.48Kohm, R1 = 7964.99Kohm, R2 = 7035.01Kohm, R3 = 2.46Kohm, R4 = 0.82Kohm, R5 = 24.08ohm, Ry = 98.63Kohm, VCC = 15.00V, VTN = 2.45V, Kn = Since 3.28(mA/V^2), C1= 6.30uF, C2=14.98uF, C3=10.96uF, calculate the lower cutoff frequency of the circuit in Hz. When making your transactions, 2 digits will be taken after the dot.
Nodal analysis is another very useful way to obtain information regarding a circuit.electric. For example, consider the 4 Ω resistor in the circuit shown below. Nodal tension to your left was named V1, while the right was named V2. Therefore, determine the values of these tensions. answer:v1 = 9,43 V ; v2 = 11,15 V
Where X = 5 and Y = 6 and. t1 = 0,5 and t2 = 0,6 Vo = 6 For the circuit of Figure, knowing that t0 = 0s and that vC (t0) = −vo. 1) Find the mathematical expression for vC(t) in the interval t0 ≤ t <t1. Give your answer in V. Answer format: A + BeCt where A, B and C are real numbers rounded to three decimal places. 2) Find the mathematical expression for iR (t) on the interval 0<t <t1. Give your answer in amps (A). Response format: D +EeCt where D and E are real numbers rounded to three decimal places. 3) Find the mathematical expression for vC (t) on the interval t1≤t <t2. Give your answer in volts (V). Response format: F + GeHt where F, G and H are real numbers rounded to three decimal places.
The circuit given below will be analyzed. β=3 α=8 a) Norton in the form I = GNV + IN of the part of the circuit seen when looking left from the A-B ends find the equivalent (parametrically equivalent). b) Let R1 = 0.5Ω, R2 = 1Ω, J = 2A, E = 1V. 10Ω to A-B ends When the Ry load impedance is connected, the power consumed at this load is Norton for the circuit. Calculate using the equivalent.
For the circuit shown in 2.24, R1=40 kΩ R2=60 kΩ, RC=750 Ω, RE=120 Ω, VCC=10 V, V0=600 mV, and β=49.Calculate iB, iC, iE, v3d, vbd, i2, i1, vab, iCC, and v13 . (Note: In thedouble subscript notation on voltage variables, the first subscript ispositive with respect to the second subscript. See P2.39.)