HW 1

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North Carolina A&T State University *

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340

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Electrical Engineering

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Apr 3, 2024

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2/7/24, 9:30 PM Assignment Print View Score: 25/25 Ppoints 100 «% Problem 01.002 Determine the current flowing through an element if the charge flow is given by the following equation. References Section Break Difficulty: Easy Problem 01.002 Learning Objective: Understand the relationship between charge and current and how to use both in a variety of applications. https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJwcmilu... 117

2/7/24, 9:30 PM Assignment Print View 1 . Award: 2.50 out of 2.50 points Problem 01.002.b - Current flowing for a quadratic time function of charge. q(t) = (202 + 11t—2) C The current i flowing through the element is (40 @t + 11 &)A. References Numeric Difficulty: Easy Response Problem Learning Objective: 01.002.b - Understand the Current flowing relationship between for a quadratic charge and current time function of and how to use both charge. in a variety of applications. Problem 01.002.b - Current flowing for a quadratic time function of charge. g = (2012 + 11t - 2) C The current j flowing through the elementis (| 40+ 2%|t+| 11+ 2%]|) A. Explanation: For the charge flow q(t) = (20 + 11t — 2) C, the current flowing through the element can be calculated as i =21 = (40t + 11) A https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJwcmilu... 2/17

2/7/24, 9:30 PM 2 . Award: 2.50 out of 2.50 points Problem 01.002.d - Current flow due to a sinusoidal time dependence of charge. q(f) = 24sin (12077t) pC The current i flowing through the element is 2880 & 1rcos (12017f) pA. References Numeric Response Problem 01.002.d - Current flow due to a sinusoidal time dependence of charge. Difficulty: Easy Learning Objective: Understand the relationship between charge and current and how to use both in a variety of applications. Assignment Print View charge. Explanation: calculated as q(f) = 24sin (1201r?) pC The current i flowing through the element is | i = Z—z = 28807 cos (1207t) pA Problem 01.002.d - Current flow due to a sinusoidal time dependence of 2880 + 2% |rcos (1201rf) pA. For the charge flow q(t) = 24sin(1201rf) pC, the current flowing through the element can be https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJIbnZpcmOubWVudCI6InByb2QiLCJpc3MiOidlenQiLCJwecmlu... 317

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2/7/24, 9:30 PM Assignment Print View 3. Award: 2.50 out of 2.50 points Problem 01.006 - Graphical computation of current The charge entering a certain element is shown in the given figure, where A = 70. Find the current at 1 ms, 6 ms, and 10 ms. g(1) (mC) A A — | | 4 6 8 N B ———— 0 The current at 1 msis 35 @ A. The currentat6 msis 0 @ A. The currentat 10 msis -17.5 @ A. References Numeric Difficulty: Easy Response Problem 01.006 Learning Objective: - Graphical Understand the computation of relationship between current charge and current and how to use both in a variety of applications. Problem 01.006 - Graphical computation of current The charge entering a certain element is shown in the given figure, where A = 70. Find the current at 1 ms, 6 ms, and 10 ms. https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJwemilu... 4/17

2/7/24, 9:30 PM Assignment Print View g(1) (mC) A A = | | E | | E 1 » 0 2 4 6 8 10 12 ;(ms) The current at 1 ms is | 35+ 2%|A. The current at 6 ms is 0%2%|A. The current at 10 ms is | -17.5+ 2% | A. Explanation: The current at 1 ms is calculated as follows: j= % _ 1m0 _ 35 A dt 2 ms The current at 6 ms is calculated as follows: i=2=0A The rate of change of charge is zero at this time. Hence, the value of current at 6 ms is zero. The current at 10 ms is calculated as follows: . dg _ —70mC __ 1= = =-175A https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJIbnZpecmOubWVudCI6InByb2QiLCJpc3MiQidlenQiLCJdwemlu. .. 5/17

2/7/24, 9:30 PM Assignment Print View 4 Award: 2.50 out of 2.50 points Problem 01.009 - Finding charge from a graph The current through an element is shown in the given figure. Determine the total charge that passed through the elementat 1 s, 3 s, and 5 s. i(A)A 10 j 5 ) 0 1 2 3 4 5 t(s) The total charge that passed through the elementatt=1sis 10 & C The total charge that passed through the element at =3 s is 22.5 @ C. The total charge that passed through the elementatt=5sis 30 @ C References Numeric Difficulty: Medium Response Problem 01.009 Learning Objective: - Finding Understand the charge froma relationship between graph charge and current and how to use both in a variety of applications. Problem 01.009 - Finding charge from a graph The current through an element is shown in the given figure. Determine the total charge that passed through the elementat 1 s, 3 s, and 5 s. https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJwcmilu... 6/17

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27124, 9:30 PM Assignment Print View i(A)A 0 1 2 3 4 5 1(s) The total charge that passed through the elementatt=1s s 10| C. The total charge that passed through the elementat t=3 s is 22.5|C. The total charge that passed through the elementatt=5s is 30| C. Explanation: The total charge that passed through the element at 1 s, 3 s, and 5 s is calculated as follows: Att=1s, g= [idt = [ 104t =10 C Att=3s, g= [Jidt=10x1+ (10 — 221) 4+ 5 x 1 =10+75+5=225C Att=5s, g= [ idt=10+10+10=30 C https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJIbnZpcm9ubWVudCI6InByb2QiLCJpc3MiOidlenQiLCdwemlu... 7/17

2/7/24, 9:30 PM Assignment Print View Problem 01.013 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS The charge entering the positive terminal of an element is g =7 sin(47f) mC, while the voltage across the element (plus to minus) is v= 2 cos(4mi) V. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. References Section Break Difficulty: Medium Problem 01.013 - Learning Objective: DEPENDENT Develop an MULTI-PART understanding of PROBLEM - power and energy ASSIGN ALL and their relationship PARTS with current and voltage. https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJwcmilu... 8/17

2/7/24, 9:30 PM 5. Assignment Print View Award: 2.50 out of 2.50 points Problem 01.013.a - Relating power, charge, and voltage Find the power delivered to the element at t = 0.3 s. The power delivered to the element at {=0.3 sis 115.15 & mW. References Numeric Difficulty: Medium Response Problem Learning Objective: 01.013.a - Develop an Relating power, understanding of charge, and power and energy voltage and their relationship with current and voltage. Problem 01.013.a - Relating power, charge, and voltage Find the power delivered to the element at t = 0.3 s. The power delivered to the elementatt=0.3sis| 115.15+ 3% | mW. Explanation: The power delivered to the element at { = 0.3 s is calculated as follows: . dg _ d(Tsin(4nt)) = — = = = 287 cos(4nt) mA p = vi = (2 cos(4mt) V) x (287 cos(4mt) mA) = 567 cos(4rt) mW Att=0.3s, p = vi = 567 cos?(47 x 0.3) mW = 115.15 mW https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJIbnZpcmOubWVudCI6InByb2QiLCJpc3MiOidlenQiLCJwecmlu... 917

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2/7/24, 9:30 PM Assignment Print View 6 . Award: 2.50 out of 2.50 points Problem 01.013.b - Integral relation between energy and power Calculate the energy delivered to the element between 0 and 0.6 s. The energy delivered to the element between 0 and 0.6 s is 54.84 @ mJ. References Numeric Difficulty: Medium Response Problem Learning Objective: 01.013.b - Develop an Integral relation understanding of between energy power and energy and power and their relationship with current and voltage. Problem 01.013.b - Integral relation between energy and power Calculate the energy delivered to the element between 0 and 0.6 s. The energy delivered to the element between 0 and 0.6 s is | 54.84 + 3% | mJ. Explanation: The energy delivered tcg éhe element between O a(?(cjl 0.6 s is calculated as follows: . 56 . W = [pdt = 567 [~ cos® (4nt) dt = <> [ [1 4 cos(8xt)] at W = 2%(0.6 + = (sin(8m x 0.6) — sin(0))) W=54.84 mJ https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJweml. .. 10/17

2/7/24, 9:30 PM Assignment Print View /. Award: 2.50 out of 2.50 points Problem 01.018 - Power absorbed in an element Find the power absorbed by each of the elements in the given figure, where /4 =15.00Aand /5 = 21.00 A. 10V 8V I§1 s = 045 —__§ 1 [ == | R P2 Pa ‘Iz " + e o + g 30V (2)p, 20-\/ P3 12v- f>0.411 The power absorbed by each of the elements is as follows: P11 = -450 & W P> = 150 & W P3 = 420 & W ps=-48 @ W ps=-72 @ W References Numeric Difficulty: Easy Response Problem 01.018 Learning Objective: - Power Begin to understand absorbed in an the volt-amp element characteristics of a variety of circuit elements. Problem 01.018 - Power absorbed in an element https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJweml. .. 11/17

27124, 9:30 PM Assignment Print View Find the power absorbed by each of the elements in the given figure, where /4 = 15.00 Aand /, = 21.00 A. The power absorbed by each of the elements is as follows: P1 =| -450 + 2%|W p2 = W p3 = w P4 = w ps=[__-72x2%|wW Explanation: The power absorbed by each element is calculated as follows: p1=vi=30V x-15.00 A=-450.000 W po=vi=10V x 15.00 A=150.000 W p3=vi=20V x21.00 A=420.000 W ps=Vvi=8V x—-6.00A=-48.000 W ps=vi=12V x-6.00 A=-72.000 W https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJIbnZpcm9ubWVudCI6InByb2QiLCJpc3MiOidlenQiLCJdweml... 1217

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2/7/24, 9:30 PM Assignment Print View Problem 01.020 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS In the figure given below, vy=30V, v, =12V, v3 =28V, and v, =28 V. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. 10 =2 A V3 | | | | + i Vv 6 A i 2 1 A Va { | | | 3A ‘ RS - + [ Vi Cfi) V, <_—'_> 3, — || 6A} 13A References Section Break Difficulty: Easy Problem 01.020 - Learning Objective: DEPENDENT Begin to understand MULTI-PART the volt-amp PROBLEM - characteristics of a ASSIGN ALL variety of circuit PARTS elements. https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJweml. .. 13/17

2/7/24, 9:30 PM 8. Assignment Print View Award: 2.50 out of 2.50 points Problem 01.020.a - Computing power with a dependent source in a circuit Calculate the power absorbed by each element in the circuit of the given figure. The power absorbed by the 30 V source is -180 & W. The power absorbed by the V,is 72 @ W. The power absorbed by the V3 is 56 @ W. The power absorbed by the V, is 28 @ W. The power absorbed by the 5/; dependent source is =30 & W. References Numeric Difficulty: Easy Response Problem Learning Objective: 01.020.a - Begin to understand Computing the volt-amp power with a characteristics of a dependent variety of circuit source in a elements. circuit https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJweml. .. Problem 01.020.a - Computing power with a dependent source in a circuit Calculate the power absorbed by each element in the circuit of the given figure. The power absorbed by the 30 V source is | -180 + 2% | W. The power absorbed by the Vo is| 72+ 2%|W. The power absorbed by the Vj is W. The power absorbed by the V4 is |T2%| W. The power absorbed by the 5/, dependent source is| -30 + 2% | W. Explanation: The power absorbed by each element is calculated as follows: P30V source =30 V X -6 A=-180 W P12V element = 12V X6 A=72W 14/17

2/7/24, 9:30 PM Assignment Print View P28 Vv element with 2 amps flowing through it =28V x2A=56W P28 Vv element with 1 amp flowing through it =28V x1A=28W Pthe 5/0 dependent source =2 X 2AX -3A=-30 W Problem 01.026 - DEPENDENT MULTI-PART PROBLEM - ASSIGN ALL PARTS A flashlight battery has a rating of 3.6 ampere-hours (Ah) over a period of 24 hours. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. References Section Break Difficulty: Medium Problem 01.026 - Learning Objective: DEPENDENT Develop an MULTI-PART understanding of PROBLEM - power and energy ASSIGN ALL and their relationship PARTS with current and voltage. https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJweml. .. 15/17

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2/7/24, 9:30 PM Assignment Print View 9. Award: 2.50 out of 2.50 points Problem 01.026.a How much constant current can it deliver over 24 hours? The current that can be delivered over 24 hours is 150 & mA. References Numeric Difficulty: Medium Response Problem Learning Objective: 01.026.a Develop an understanding of power and energy and their relationship with current and voltage. Problem 01.026.a How much constant current can it deliver over 24 hours? The current that can be delivered over 24 hoursis| 150 £ 2% | mA. Explanation: The amount of current that can be delivered over 24 hours is ; — (3.2641;.11) (10(10;1,4) — 150 mA https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJweml. .. 16/17

2/7/24, 9:30 PM 10. Assignment Print View Award: 2.50 out of 2.50 points Problem 01.026.b - Relating power to current rating and voltage. How much constant power can it deliver over 24 hours if its terminal voltage is 20 V? Constant power that can be delivered over 24 hours if its terminal voltage is 20 V is 3000 & mW. References Numeric Difficulty: Medium Response Problem Learning Objective: 01.026.b - Develop an Relating power understanding of to current power and energy rating and and their relationship voltage. with current and voltage. Problem 01.026.b - Relating power to current rating and voltage. How much constant power can it deliver over 24 hours if its terminal voltage is 20 V? Constant power that can be delivered over 24 hours if its terminal voltage is 20 Vis| 3000 + 2% | mW. Explanation: The constant power can be calculated from the following equation: p=vi=(20V x 015 &) (L0EW) _ 3000 mW https://ezto.mheducation.com/api/caa/activity/C15Print?jwt=eyJhbGciOiJSUzI1NiJ9.eyJibnZpcmOubWVudCI6InByb2QiLCJpc3MiOiJlenQiLCJweml. .. 17/17

HW 1

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