Follow the instructions. Typewritten for an upvote. No upvote for handwritten. PLEASE SKIP IF YOU HAVE ALREADY DONE THIS. Thank you Table 5-1.DescriptionMeasuredE-Audio Generator Output voltage10VppER Voltage across resistor R7.4837021 VppEc Voltage across capacitor C6.6208014 VppITotal currentI+= ER/R0.416 mAZ Total impedanceZ+=E+/+Z₁ = 24054.524Ω8-Phase angle between E, and ER 1.1514986ms/10ms x 360° =41.4539496⁰0 Phase angle between E, and E.1.3284015ms/10ms x 360º =47.822454⁰5.2. What are the possible causes of error in the measurement of angle Or and 8?%diffComputed10 Vpp0%0.11%ER=I, XR7.491541564 VppEc= ITX Xc0.05%note X₂= 1/(2TFC)6.623968221 VppI+= E-/Z+0.23%I+= 0.41696198 mAZ-= √(R²+X²)0.11%Xc = 15915.49431Z₁ = 24027.12965 Q8,= tan ¹(X/R)0.07%0,=41.483⁰8 = tan ¹(R/X) = 90° - eT 1.44%0 = 48.517⁰[

Capacitive reactance inversely porotional to frequency Inductive reactance directly proportional to…

Table 5-1. Description Measured Computed E-Audio Generator Output voltage 10Vpp 10 Vpp ER Voltage across resistor R 7.4837021 Vpp ER=I, XR 7.491541564 Vpp Ec Voltage across capacitor C 6.6208014 Vpp Ec= ITX Xc note X₂= 1/(2TFC) 6.623968221 Vpp I Total current I+= E-/Z+ I+= ER/R 0.416 mA I+= 0.41696198 mA Z Total impedance Z+=E+/+ Z-= √(R²+X²) Z₁ = 24054.524 Ω Xc = 15915.49431 Z₁ = 24027.12965 Q 8- 8,= tan ¹(X/R) Phase angle between E, and ER 1.1514986ms/10 ms x 360° = 41.4539496⁰ 0,=41.483⁰ 0 Phase angle between E, and E. 1.3284015ms/10 ms x 360º = 47.822454⁰ 8 = tan ¹(R/X) = 90° - eT 0 = 48.517⁰ 5.2. What are the possible causes of error in the measurement of angle Or and 8? %diff 0% 0.11% 0.05% 0.23% 0.11% 0.07% 1.44%

Table 5-1. Description Measured Computed E-Audio Generator Output voltage 10Vpp 10 Vpp ER Voltage across resistor R 7.4837021 Vpp ER=I, XR 7.491541564 Vpp Ec Voltage across capacitor C 6.6208014 Vpp Ec= ITX Xc note X₂= 1/(2TFC) 6.623968221 Vpp I Total current I+= E-/Z+ I+= ER/R 0.416 mA I+= 0.41696198 mA Z Total impedance Z+=E+/+ Z-= √(R²+X²) Z₁ = 24054.524 Ω Xc = 15915.49431 Z₁ = 24027.12965 Q 8- 8,= tan ¹(X/R) Phase angle between E, and ER 1.1514986ms/10 ms x 360° = 41.4539496⁰ 0,=41.483⁰ 0 Phase angle between E, and E. 1.3284015ms/10 ms x 360º = 47.822454⁰ 8 = tan ¹(R/X) = 90° - eT 0 = 48.517⁰ 5.2. What are the possible causes of error in the measurement of angle Or and 8? %diff 0% 0.11% 0.05% 0.23% 0.11% 0.07% 1.44%

Power System Analysis and Design (MindTap Course List)

6th Edition

ISBN:9781305632134

Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Publisher:J. Duncan Glover, Thomas Overbye, Mulukutla S. Sarma

Chapter6: Power Flows

Section: Chapter Questions

Problem 6.61P

See similar textbooks

Similar questions

estimate the value of equivalent impedance
1. If ?1 = 30sin (?? + 100) and ?2 = 20sin (?? + 500), which of these statements is true: (a) ?1 leads ?2 (b) ?2 leads ?1 (c) ?1 lags ?2 (d) ?2 lags ?1 (e) ?1 and ?2are in phase 2.A series RC circuit has |??| = 12 ? and |??| = 5 ?. The magnitude of the supply voltage is: (a) -7 V (b) 7 V (c) 13 V (d) 17 V
A cable has inductance of 0.22 mH per km and capacitance of 0.202 microF per km. The surge impedance of the cable is ?
In series reactive power compensation, the series capacitors are not used to increase the capabilities of transmission lines. Select one: True False
help me calculate the Rn, In or Isc, and current at Zc using NORTON’S THEOREM
A certain industrial load has an impedance of 0.6 angle 30 ohms at a frequency of 60 Hz. What capacitorconnected in parallel with this load causes the angle of the total impedance to decrease to 15º ? Also if the load voltage is 120 V, what is the decrease in line current produced by adding the capacitor ?
Which termination would result to a phase change of 180 degrees at the load if Zo = 100 ohm? a. 50 ohm b. short circuit c. 75 ohm d. all of these What is the purpose of impedance matching? a. maximum return loss b. maximum SWR c. maximum load power d. maximum reflection A short piece of transmission line that may be open or shorted and used for impedance matching purposes. a. converter b. decoder c. stub d. transformer
An8pole50Hz,threephaseinductionmotorrunsataspeedof720rpm whendeliveringfullloadtorque.Itsstandstillrotorresistanceandreactance are0.1ohm and0.6ohm perphaserespectively.Anadditionalresistanceof 0.5ohm perphaseisinsertedintherotorcircuittocontrolthespeed. Calculatethespeedatwhichthemotorrunsnow withfullloadtorque remainingthesame.
Athreephase,50Hz,4poleslipringinductionmotorrunsat1455rpmonfull load.Therotorresistanceperphaseonstandstillis0.2ohm.Itisrequiredto reducethespeedby10%byinsertingtheseriesresistanceintherotorcircuit. Determinetherotorresistanceperphaserequired.
In a single-phase transmission line circuit, three 0.25 It consists of a conductor with a radius of cm (a, b, c). The return circuit likewise consists of three parts of the same radius. consists of two conductors (d, e, f). of conductors placement is given. of each conductor Calculate the inductance and the total inductance.
A single circuit three phase 50 Hz, 55 km transmission line consists of three conductors arranged in triangle. d1=5m, d2=9 m and d3=19 m. The radius is 0.7 cm. Calculate a) GMD b) Inductance per km c) Total Inductance and d) Inductive reactance.
The power factor angle of a purely inductive circuit is 0 degrees Select one: True False