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Ayear in the modern gregorian calendar consists of 365 days. in reality, the earth takes longer to rotate around the sun. to account for the difference in time, every 4 years, a leap year takes place. a leap year is when a year has 366 days: an extra day, february 29th. the requirements for a given year to be a leap year are: the year must be divisible by 4if the year is a century year (1700, 1800, the year must be evenly divisible by 400some example leap years are 1600, 1712, and 2016.write a program that takes in a year and determines whether that year is a leap year.ex: if the input is 1712, the output is: 1712 is a leap year.ex: if the input is 1913, the output is: 1913 is not a leap year.

Write a computer program to calculate the three-phase fault current for a fault at f in figure 1.16, with the network normal, and with one line at a time removedproblems 1.1 write a computer program to calculate the three-phase fault current for a fault at f in figure 1.16, with the network normal, and with one line at a time removed 20 power system relaying from service. the positive-sequence impedance data are given in the accompanying table. use the commonly made assumption that all prefault resistance values are (1.0+j0.0) pu, and neglect all resistance values. calculate the contribution to the fault flowing through the cb b and the voltage at that bus. for each calculated case, consider the two possibilities: cb b2 closed or open. the latter is known as the stub-end fault í• figure 1.16 problem 1.1 system data for figure 1.16 from to positive sequence 0.0+j0.1 0.05j0.15 0.04 j0.2 0.01 jo.i 0.015 + j0.15 0.01 j0.19 0.01 +j0.19 0.03+j0.1 0.0+j0.08 6 6 6 from service. the positive-sequence impedance data are given in the accompanying table. use the commonly made assumption that all prefault resistance values are (1.0 + j 0.0) pu, and neglect all resistance values. calculate the contribution to the fault flowing through the cb b1, and the voltage at that bus. for each calculated case, consider the two possibilities: cb b2 closed or open. the latter is known as the “stub-end” fault.

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