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
Explanation:
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Answer:
Hello there, the question is not complete, but not to worry you can check the explanation section to check how you can solve a similar question or to be be able to solve the exact question directly.
Explanation:
The flow of electrons is what is known or refer to as Current. When energy is used on a nuclei, the electrons are forced to move from one position to the other. The direction of flow of electron is from the negative terminal which then moves to the positive terminal.
Therefore, it can be said that the positive charge determines the direction of electron flow. The starting point is the negative terminal, in which it will now move in the direction in which the positive terminal is.
This question is incomplete, its missing an image which will be uploaded along this Answer.
Answer:
the normal component of force F_n is F((√(r²-s²)) / r)
the tangential component of force F_t is F(s/r)
Explanation:
Given the data in the image;
from the free body diagram, we write the expression for ∅
sin∅ = s/r
cos∅ = (√(r²-s²)) / r
now expression for normal component of force is;
F_n = Fcos∅
we substitute
F_n = F((√(r²-s²)) / r)
Therefore, the normal component of force F_n is F((√(r²-s²)) / r)
Also for force F_t
F_t = Fsin∅
we substitute
F_t = F(s/r)
Therefore, the tangential component of force F_t is F(s/r)
