Answer:
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
Explanation:
The peak voltage after the 6 to 1 step down is . Then, the peak voltage of the rectified output is V_{d}[/tex] and according to the statement, the diodes can be modeled to be . Then, the peak voltage in the load is .
The upper diode conduces in the odd half cycles. The lower diode conduces in the even half cycles.
The average output voltage is calculated as:
The average current in the load is calculated as:
Answer:
False
Explanation:
cost Evaluation
assembly of equipment and materials used
Answer:
#Selling vehicles
import locale
locale.setlocale( locale.LC_ALL, 'en_CA.UTF-8' )
#Declaration of variables
total_pay=0
total_sales=0
#Selling details
for i in range(0,2):
#Ask the type of sell
type=input("Enter the type of the car you sold(used/new)? ")
#Check error
while(type.upper()!="USED" and type.upper()!="NEW"):
print('ERROR!!!Should be used or new!!Please Re-enter')
type=input("Enter the type of the car you sold(used/new)? ")
#Input price of the car
price=float(input("Enter the price of the car: "))
#Calculations
if(type.upper()=="NEW"):
total_pay+=1500
else:
total_pay+=price*.05
total_sales+=price
#Display results
print('Total Pay of the sale person = ',locale.currency(total_pay))
print('Total Sales = ',locale.currency(total_sales))
Answer:
T=833.8 °C
Explanation:
Given that
m= 2 kg
T₁=200 °C
time ,t= 10 min = 600 s
Work input = 1 KW
Work input = 1 x 600 KJ=600 KJ
Heat input = 0.5 KW
Q= 05 x 600 = 300 KJ
Gas is ideal gas.
We know that for ideal gas internal energy change given as
ΔU= m Cv ΔT
For air Cv= 0.71 KJ/kgK
From first law of thermodynamics
Q = ΔU +W
Heat input taken as positive and work in put taken as negative.
300 KJ = - 600 KJ + ΔU
ΔU = 900 KJ
ΔU= m Cv ΔT
900 KJ = 2 x 0.71 x (T- 200 )
T=833.8 °C
So the final temperature is T=833.8 °C