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Ans: ( 0.75A) 240 幸る2 寺 キ42 Fig. c4) Q2 / In the circuit shown in fig.(2) calculate (I) using current divider rule. Ans: (0.25A) 13 24U 222 Fy-(2)

Ans: ( 0.75A) 240 幸る2 寺 キ42 Fig. c4) Q2 / In the circuit shown in fig.(2) calculate (I) using current divider rule. Ans: (0.25A) 13 24U 222 Fy-(2)

Introductory Circuit Analysis (13th Edition)
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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Tutorial Sheet No.1 Q/A hydro-electric generating plant is supplied from a reservoir of capacity 20000000m with a head of 200m. The hydraulic efficiency of the plant is 0.8 and the electric efficiency is 0.9. What is the total available energy? Ans: [7.85GWH] Q5/ The area of the reservoir in problem 4 is 3.0km². The plant supplies a load of 12MW for 3h. Calculate the fall in the level of the reservoir during this period. Ans: [30.5m] Q6/ The reservoir in problem 4 is supplied by a river at the rate of 2m³/s. Assuming constant head and efficiencies, what does this flow represent in terms of megawatts, megawatt hours per-day and gigawatt hours per annum? Ans: [2.825MW, 67.8MWh/d ,24.408GWH/a]
Transcribed Image Text:Tutorial Sheet No.1 Q/A hydro-electric generating plant is supplied from a reservoir of capacity 20000000m with a head of 200m. The hydraulic efficiency of the plant is 0.8 and the electric efficiency is 0.9. What is the total available energy? Ans: [7.85GWH] Q5/ The area of the reservoir in problem 4 is 3.0km². The plant supplies a load of 12MW for 3h. Calculate the fall in the level of the reservoir during this period. Ans: [30.5m] Q6/ The reservoir in problem 4 is supplied by a river at the rate of 2m³/s. Assuming constant head and efficiencies, what does this flow represent in terms of megawatts, megawatt hours per-day and gigawatt hours per annum? Ans: [2.825MW, 67.8MWh/d ,24.408GWH/a]
1,4 Tutorial Sheet No.3 Q1/In the circuit shown in fig.(1) calculate (I) using current divider rule. Ans: (0.75A) 240 Fig. C4) Q2 / In the circuit shown in fig.(2) calculate (I) using current divider rule. Ans: (0.25A) 132 24U 222 * 52 Fiy-(2) Q3 / calculate the voltage across AB in the network shown in fig.(3) and indicates the polarity of the voltage. Ans: (B is 2V above A) A B. Fig. (3) Q4 / Determine the branch current in the network of fig.(4) when the value of each branch resistance is 12. Ans: (AB=BC=6.25A BD=0 AD=DC=1.25A ;CA=7.5A) 5V A ir lov Fig. (4)
Transcribed Image Text:1,4 Tutorial Sheet No.3 Q1/In the circuit shown in fig.(1) calculate (I) using current divider rule. Ans: (0.75A) 240 Fig. C4) Q2 / In the circuit shown in fig.(2) calculate (I) using current divider rule. Ans: (0.25A) 132 24U 222 * 52 Fiy-(2) Q3 / calculate the voltage across AB in the network shown in fig.(3) and indicates the polarity of the voltage. Ans: (B is 2V above A) A B. Fig. (3) Q4 / Determine the branch current in the network of fig.(4) when the value of each branch resistance is 12. Ans: (AB=BC=6.25A BD=0 AD=DC=1.25A ;CA=7.5A) 5V A ir lov Fig. (4)
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