6. (15 Points) Thévenin Equivalent: Obtain the Thévenin equivalent seen by the (2 - j) 2 impedanceof Fig. 10.69, and employ it to determine the current 1₁.1.5/24° AFIGURE 10.69j2Q2(2-j) QRef.4 /10° Ω2/38° A

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Answered: (15 Points) Thévenin Equivalent: Obtain…

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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2. Express the following functions as a cosine: a. 10 sin(wt +30°) b. -40 sin(wt - 70°) c. 25 sin(wt 15°) d. 5 sin(wt +230°) - 3. For the two signals Va(t) and Va(t) below, draw the phasor diagram for both signals, then determine which one leads and by how much: Va(t) = 12 cos(4t-80°) Volts Vb (t) = 10 sin(4t + 20°) Volts complex numbers, express the results in polar form. 4. Evaluate the following a. 3 + 4j b. -11j c. -12 + 15j d. 5 - 4j 5. Find the phasors corresponding to each of the following signals: a. v(t) = 12 cos (4t- 200°) Volts b. i(t) = -50 sin(100t +30°) Amps c. i(t) = -45 cos(6t+ 40°) Amps d. v(t) = 8 sin (200t - 170°) Volts
4) For the following pairs of sinusoids, determine which one leads and by how much. a) v(t) = 10 cos(4t – 60°) and i(t) = 4 sin(4t + 50°) b) vị (t) = 4 cos(377t + 10°) and v2 (t) = -20 cos 377t c) i(t) = – sin(wt + 30°) and v(t) = 2 sin(wt + 10°) d) x(t) = 13 cos 2t + 5 sin 2t and y(t) = 15 cos(2t – 11.8°)
A circuit branch is found to have an equivalent impedance of Z=120−j45Ω. It is given that this impedance is a series combination of a 120Ω resistor and another passive component. If the circuit has a sinusoidal voltage source whose frequency is 120 Hz. What is the value of the component in series with the 120Ω resistor? (Note: Express the value assuming a metric scaling of μ(10-6). Provide your answer in decimal form, up to 3 decimal places.)
Given the circuit shown below while assuming a sinusoidal steady state, 50 0 250 0 2 A 10 Vi -j50 0 +j 50 0 If we apply a source Transformation the voltage source becomes equal to: Select one: O a. 10- j10 V O b. 25- j25 V OC 50-j50 V Od. None of these
2. What is the value of a sinusoidal voltage at 5 us (micro second) from the positive going zero crossing when V, = 20V and f = 50KHZ ? %3D 3. A sinusoidal voltage is applied to the circuit below. Determine the following: Ip, Ims, lavg, Ipp-
The signal V1=100 cos(500t - 30°) and the signal V2=20 sin (500t +45°) compared in phase would be a) V1 15° ahead of V2 b) V1 75° behind V2 c) V2 15° ahead of V1 d) None of the above
Convert the circuit below from the time-domain to the phasor-domain. R1 R2 70 130 Vs = 3 sin (25t + 110°) L 160mH C :2mF a) What is the magnitude of Vs, the voltage source, in the phasor domain? b) What is the phase of Vs, the voltage source, in the phasor domain? c) What is the impedance of resistor R1, Zr1? d) What is the impedance of resistor R2, Zr2? e) What is the impedance of the capacitor, Zc? f) What is the impedance of the inductor, Z1?
For the circuit shown, the RMS value of lo is. 4KVA 0.8 pf leading 120430°(± 2.4KVA 1KW V rms 0.6 pf lagging 500 VAR Select one: O a. 94 A ms. O b. 2.7 Amms. O c. 47 A rms. O d. 50 A rms.
five infinite-impedance volunteers,calibrated to read rms values,are connected as shown below. let R= 200, L=0.400H,C=6.0 Mf, and V=30.0 v. what's the reading of the voltmeter's V1,V2,V3,V4 and V5 if W= 200 rad/s
Find the equivalent impedance Zeq seen by the source when Vs = 4 cos (2t) v, C = 1 F, R = 3 2 and L = 0.5 H. (Give angles in degrees and round final answers to two decimal places.) Calculate the voltage across each component - capacitor, resistor and inductor. (Give angles in degrees and round final answers to two decimal places.) Vs R The equivalent impedance seen by the source is Q. The voltage across the capacitor is The voltage across the resistor is The voltage across the inductor is at an angle of at an angle of at an angle of + V. V. V. Q= at an angle of
Given the circuit shown below while assuming a sinusoidal steady state, 50 0 250 0 2 A -/50 0 */ 50 0 If we apply a source Transformation the voltage source becomes equal to: Select one: O a. None of these O b. 100+ j100 V Oc 100-j100 V O d. 10-j10 V
5. A voltage v(t) = 230 sin ot when applied to a circuit supplies a current given by the expression i(t) = 8cosot +8cos(@t + 60°) +6sin(t +30°)+2sin(@t – 30°)A (a) Using phasor diagram or otherwise, find the amplitude or maximum value of the 14. A oh current. [Ans. 114] (Ans 90 07 15. Th (b) Find the phase angle between the voltage and the current

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(15 Points) Thévenin Equivalent: Obtain the Thévenin equivalent seen by the (2 − j) impedance of Fig. 10.69, and employ it to determine the current l₁. 1.5/24° A FIGURE 10.69 j2 s2 (2-j) 22 Ref. 4/10° 92 2/38° A