III. PROBLEM-SOLVING DIRECTION: Solve the following problem manually on a clean sheet of paper. Box your final answer with corresponding signature. Make sure you round the final answer to four decimal places, magnitude only and label it with the correct units. Avoid too much ERASURE. A. A circuit consisting of a coil with inductance 10 mH and resistance 20 ohms is connect in series with a capacitor and a generator with an rms voltage of 120 V. Find: (10 points) a. The value of the capacitance that will cause the circuit to be in resonance at 15 kHz b. The current through the coil at resonance c. The Q of the circuit SOLUTION a. Solving the capacitance of the circuit L = 10 mH; R = 20 Ohms; Vs=120 V; fr = 15 kHz 1 ππ = 2π√πΏπΆ 1 15π₯103 = 2π√(10π₯10−3 )πΆ 1 √(10π₯10−3 )πΆ = 2π(15π₯103 ) 2 (√(10π₯10−3 )πΆ) = (1.0610π10−5 )2 (1.1258π10^ − 10) (10π₯10−3 ) πΆ = 11.258π10−9 π πͺ = ππ. πππ ππ b. Solving the current through the coil at resonance Consider the visualization of the circuit on the figure below πΆ= ππ = 120π Converting the circuit to the domain of frequency π€π = 2πππ π€π = 2π(15π₯103 ) π€π = 94.2478 ππ»π§ ππΏ = ππ€πΏ ππΏ = π(94.2478π₯103 )(100π₯10−3 ) ππΏ = π9424.7780 πβππ ππ 9425 πβππ 1 ππΆ = ππ€πΆ 1 ππΆ = 3 π(94.2478π₯10 )(1.1258π₯10−9 ) ππΆ = −π9424.6999 πβππ ππ − π9425 πβππ ππ = π + πππΏ − πππΆ ππ = 20 + π9425 − π9425 ππ = 20 πβππ ππ πΌπ = ππ 120 π πΌπ = 20 πβππ π°π» = π π¨ c. Solving Quality Factor 1 π€π πΏ π= ππ π€π πΆπ π π= (94.2478π₯103 )(10π₯10−3 ) 1 ππ (94.2478π₯103 )(11.258π₯10−9 )(20) 20 πΈ = ππ. ππππ ππ ππ. ππππ B. Two wattmeters are connected to measure the input power to a balanced three-phase load. If the wattmeter readings are 9.3 kW and 5.4 kW, determine (5 points) a. the total output power, and b. the load power factor SOLUTION a. Total output power ππ = πππ 1 + πππ 2 ππ = 9300π + 5400π π·π» = πππππ πΎ ππ ππ. πππΎ b. Load Power Factor √3(πππ 1 − πππ 2 ) πππ 1 + πππ 2 √3(9300π − 5400π) π‘πππ = 9300π + 5400π 13 π = tan−1( √3) 49 π = 24.6799 ππ = πππ π ππ = πππ 24.6799 ππ = π. ππππ π‘πππ = C. Solution: https://drive.google.com/file/d/114Za78KVClFLoIe9AcBFAEsmfgC54_7/view?usp=sharing D. Solution: https://drive.google.com/file/d/1EIOzjOTAvw5fNP2xWLPq1zzE1HGuHlRR/view? usp=sharing E. Solution:https://drive.google.com/file/d/14Pm_B8ApX-hV66lWte9gGvzoBB_l4UL/view?usp=sharing VI. NETWORK ANALYSIS DIRECTION: Solve the following problem manually on a clean sheet of paper. Box your final answer with the corresponding signature. Make sure you round the final answer to four decimal places, polar form, and label it with the correct units. Avoid too much ERASURE. 24. For the three phase circuit below IbB = 30∠ 60° A and VBC = 220∠ 10° V. Find the Van, VAB, IAC and Z. (10 points) SOLUTION: ` πππ = πππ √3∠−90° = 220∠10° √3∠−90° ; πππ = ππ΅πΆ π½ππ = πππ∠πππ° π½ Since balance system, we can relate them with the positive sequence of the network. ππ΄π΅ = ππ΅πΆ ∠π + 120 = 220∠10 + 120 = 220∠130° π π½π¨π© = πππ∠πππ° π½ ππ΄πΆ = ππ΄πΆ ∠π − 120 = 220∠10 − 120 = 220∠ − 110° π If the value of IbB is 30∠60° A , then by the identity of the balance system πΌππ΄ = πΌππ΅ ∠π + 120 = 30∠60 + 120 = 30∠180° π΄ πΌππΆ = ππ΅πΆ ∠π − 120 = 30∠60 − 120 = 30∠ − 60° π΄ Converting phase current to line current. πΌπ΄π΅ = πΌππ΄ √3∠−30° = 30∠180° √3∠−30° = 17.3205∠ − 150° π΄ ππ πΌπ΄π΅ = 17.3205∠210° π΄ πΌπ΅πΆ = πΌππ΅ √3∠−30° = 30∠60° √3∠−30° = 17.3205∠90° π΄ ππ πΌπ΅πΆ = 17.3205∠ − 270° π΄ πΌπΆπ΄ = πΌππΆ √3∠−30° = 30∠−60° √3∠−30° = 17.3205∠ − 30° π΄ ππ πΌπΆπ΄ = 17.3205∠330° π΄ πΌπ΄πΆ = −πΌπΆπ΄ = −1(17.3205∠ − 30° π΄) π°π¨πͺ = (ππ. ππππ∠πππ° π¨) = (ππ. ππππ∠ − πππ° π¨) Use the gained value to obtain the impedance of any phase since they are a balance system. ππ΅πΆ π= πΌπ΅πΆ 220∠10° π π= 17.3205∠90° π΄ π = ππ. ππππ∠ − ππ° π¨ 25. Find the Vx in the network below: (10 points) SOLUTION: For a much more straightforward approach, convert the third loop by source transformation, and the circuit will look like the figure below. ππππ€ = πΌ(π) = 2∠0°π΄(−π1πβππ ) = −π2π 2Ω 8∠ππ° I1 2Ω j4 j4 -j1 I2 -j2V Create two equations for the created loops I1 and I2. For Loop 1 −8∠30° + (2 + π4)πΌ1 − ππΌ2 = 0 (2 + π4)πΌ1 − ππΌ2 = 8∠30° → πππ’ππ‘πππ 1 For Loop 2 (2 + π4 − π)πΌ2 − ππΌ1 = 0 → πππ’ππ‘πππ 2 Writing equation 2 in terms of I1 (2 + π4 − π)πΌ2 − ππΌ1 = 0 (2 + π4 − π)πΌ2 = πΌ1 π (3 − π2)πΌ2 − 2 = πΌ1 → πππ’ππ‘πππ 3 Substitute equation 3 to equation 1 to gain I2 (2 + π4)πΌ1 − ππΌ2 = 8∠30° (2 + π4)[(3 − π2)πΌ2 − 2] − ππΌ2 = 8∠30° (14 + π8)πΌ2 − 4 − π8 − ππΌ2 = 8∠30° (14 + π7)πΌ2 − 4 − π8 = 8∠30° 8∠30° + 4 + π8 πΌ2 = 14 + π7 πΌ2 = 1.0369∠21.1113° π΄ Solving Vx ππ₯ = 2πΌ2 ππ₯ = 2πβππ (1.0369∠21.1113° π΄) π½π = π. ππππ∠ππ. ππππ° π½ 26. A three-phase supply, with the line voltage 240 V rms positively phased, has an unbalanced delta-connected load as shown below. Find the phase currents and the total complex power. (10 points) SOLUTION: The system is unbalanced, so computation of the phase current will be done individually. Assuming VAB is the reference voltage. Calculating the phase current IAB ππ΄π΅ πΌπ΄π΅ = ππ΄π΅ 240∠0° π πΌπ΄π΅ = 25 πβππ π°π¨π© = π. π∠ − ππ° π¨ Calculating the phase current IBC ππ΅πΆ πΌπ΅πΆ = ππ΅πΆ 240∠ − 120° π πΌπ΅πΆ = 30∠30° πβππ π°π©πͺ = π∠ − πππ° π¨ Calculating the phase current ICA ππΆπ΄ πΌπΆπ΄ = ππΆπ΄ 240∠120° π πΌπΆπ΄ = 40 πβππ π°πͺπ¨ = π∠πππ° π¨ Calculating the power used at phase A. ππ΄ = |πΌπ΄π΅ |2 ππ΄π΅ ππ΄ = |9.6|2 (π25) ππ΄ = π2304ππ΄ Calculating the power used at phase B. ππ΅ = |πΌπ΅πΆ |2 ππ΅πΆ ππ΅ = |8|2 (30∠30° ) ππ΅ = 1662.7688 + π960 ππ΄ Calculating the power used at phase C. ππΆ = |πΌπΆπ΄ |2 ππΆπ΄ ππΆ = |6|2 (40 ππΆ = 1440ππ΄ Solving total complex power ππ = ππ΄ + ππ΅ + ππΆ ππ = π2304ππ΄ + 1662.7688 + π960 ππ΄ + 1440ππ΄ ππ = 3102.7688 + π3264 ππ΄ πΊπ» = ππππ. ππππ∠ππ. ππππ° π½π¨ 27. Three zones of a single phase circuit are identified in the figure. The zones are connected by the treansformers T1 and T2, whose rating shown below. ZLOAD=0.5+j0.2 Using base values of 30 kVA and 240 volts in zone 1, draw the pu circuit and determine the pu impedance and pu source voltage. Then calculate the load current in pu ans in amperes. (15 points) SOLUTION: Sbase=30 kVA Vbase per region 240π ππ§πππ1 = 240ππ₯ = 240π 240π 480π ππ§πππ2 = 240ππ₯ = 480π 240π 460π ππ§πππ2 = 480ππ₯ = 120π 115π Per unit computation Generator 1 ππππ£ππ 240π ππΊ1 = = = 1.0 ππ’ ππππ π 240π Transformer 1 ππππ£ππ 2 ππππ π 240π 2 30πππ΄ π₯π1 = π₯πππ£ππ ( ) ( ) = π0.1 ( ) ( ) ππππ π ππππ£ππ 240π 30πππ΄ π₯π1 = π0.1 ππ’ Transformer 2 ππππ£ππ 2 ππππ π 240π 2 30πππ΄ π₯π2 = π₯πππ£ππ ( ) ( ) = π0.1 ( ) ( ) ππππ π ππππ£ππ 240π 20πππ΄ π₯π2 = π0.1378 ππ’ Line 2 ππππ£ππ ππππ£ππ ππππ π π2 π₯πΏπΌππΈ = ; ππππ π = ; π₯ππππ = 2 = 480π 2 ππππ π ππππ π ππππ π 30πππ ππππ π π₯πΏπΌππΈ = π0.2604 ππ’ Load 2 ππππ£ππ ππππ£ππ ππππ π 0.5 + π0.2 π₯πΏππ΄π· = ; ππππ π = ; π₯ππππ = 2 = 120π 2 ππππ π ππππ π ππππ π 30πππ ππππ π π₯πΏππ΄π· = 1.0417 + π0.4167 ππ’ REACTANCE DIAGRAM IL Solving the current of the system through KVL. 1.0 − πΌπΏ (π0.1 + π0.2604 + π0.1377 + 1.0417 + π0.4167 = 0 1.0 πΌπΏ = 1.0417 + π0.9148 πΌπΏ = 0.7213∠ − 41.2890° Converting per unit value to the actual value. πΌπ΄ππ‘π’ππ = πΌππ’ ∗ πΌπππ π ππ π π‘βπ π£πππ‘πππ πππ π π€βπππ π‘βπ ππππ ππ πππππ‘ππ π€βππβ ππ π§πππ 3 ππππ π 30ππ£π΄ πΌπππ π = = = 250π΄ ππππ π 120π πΌπ΄ππ‘π’ππ = 0.7213∠ − 41.2890° ∗ 250π΄ π°π¨πππππ = πππ. ππππ∠ − ππ. ππππ°π¨ [Total Points = 60] Prepared by: Engr. Jefrey Jay S. Claus Instructor, EE Dept. Date: May 20, 2022 Engr. Analyn Cueto Instructor, EE Dept. Date: May 25,2022
