which of the following processes would be appropriate for cutting a narrow slot, less than 0.015 inch wide, in a 3/8- inch thick
1 answer:
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Answer:
A) 11.1 ms
B) 5.62 Ω
Explanation:
L ( inductance ) = 10 mH
Vcc = 14V
<u>A) determine the required on time of the switch such that the peak energy stored in the inductor is 1.2J </u>
first calculate for the current ( i ) using the equation for energy stored in an inductor hence
i = ----- ( 1 )
where : W = 1.2j , L = 10 mH
Input values into equation 1
i = 15.49 A
Now determine the time required with expression below
i( t ) = 15.49 A
L = 10 mH, Vcc = 14
hence the time required ( T-on ) = 11.1 ms
attached below is detailed solution
B) <u>select the value of R such that switching cycle can be repeated every 20 ms </u>
using the expression below
τ = ---- ( 2 )
but first we will determine the value of τ
τ = t-off / 5 time constants
= (20 - 11.1 ) / 5 = 1.78 ms
Back to equation 2
R = L / τ
= (10 * 10^-3) / (1.78 * 10^-3)
= 5.62 Ω
Answer:
1. The net power developed=9370.773KW
2. Thermal Efficiency= 0.058
Explanation
Check attachment
