9. What is a whip check?
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Answer:
(b) By including a statement that he or she is licensed by the Board for Professional Engineers and Land Surveyors immediately above the signature line in at least 12 point type on all contracts for services
Explanation:
A PE(professional engineer) licensee must show that he is licensed in order to show and ensure public safety as he is qualified for the job he is handling. The California regulations on professional engineers holds that all professional engineers must be licensed by the board of professional engineers and Land surveyors in order to operate legally as an engineer. The engineer may show licensure through the following options:
The engineer might provide statement to each client to show he is licensed which would then be signed by the client
The engineer may choose to post a wall certificate in his work premises to show he is licensed
The engineer may choose to include a statement of license in a letterhead or contract document which must be above the client's signature line and not less than 12 point type
Answer:
a) the output voltage is 50 V
b)
- the average inductor current is 10 A
- the maximum inductor current is 13 A
- the maximum inductor current is 7 A
c) the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components is 4 A
Explanation:
Given the data in the question;
a) the output voltage
V₀ = V/( 1 - D )
given that; V = 20 V, D = 0.6
we substitute
V₀ = 20 / ( 1 - 0.6 )
V₀ = 20 / 0.4
V₀ = 50 V
Therefore, the output voltage is 50 V
b)
- the average inductor current
= V
/ ( 1 - D )²R
given that R = 12.5 Ω, V = 20 V, D = 0.6
we substitute
= 20 / (( 1 - 0.6 )² × 12.5)
= 20 / (( 0.4)² × 12.5)
= 20 / ( 0.16 × 12.5 )
= 20 / 2
= 10 A
Therefore, the average inductor current is 10 A
- the maximum inductor current
= [V
/ ( 1 - D )²R] + [ V
given that, R = 12.5 Ω, V = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
= [20 / (( 1 - 0.6 )² × 12.5)] + [ (20 × 0.6 × 5) / (2 × 10) ]
= [20 / 2 ] + [ 60 / 20 ]
= 10 + 3
= 13 A
Therefore, the maximum inductor current is 13 A
- The minimum inductor current
= [V
/ ( 1 - D )²R] - [ V
given that, R = 12.5 Ω, V = 20 V, D = 0.6, L = 10 μH, T = 1/200 kHz = 5 hz
we substitute
= [20 / (( 1 - 0.6 )² × 12.5)] - [ (20 × 0.6 × 5) / (2 × 10) ]
= [20 / 2 ] -[ 60 / 20 ]
= 10 - 3
= 7 A
Therefore, the maximum inductor current is 7 A
c) the output voltage ripple
ΔV₀/V₀ = D/RCf
given that; R = 12.5 Ω, C = 40 μF = 40 × 10⁻⁶ F, D = 0.6, f = 200 Khz = 2 × 10⁵ Hz
we substitute
ΔV₀/V₀ = 0.6 / (12.5 × (40 × 10⁻⁶) × (2 × 10⁵) )
ΔV₀/V₀ = 0.6 / 100
ΔV₀/V₀ = 0.006 or 0.6%V₀
Therefore, the output voltage ripple is 0.006 or 0.6%V₀
d) the average current in the diode under ideal components;
under ideal components; diode current = output current
hence the diode current will be;
= V₀/R
as V₀ = 50 V and R = 12.5 Ω
we substitute
= 50 / 12.5
= 4 A
Therefore, the average current in the diode under ideal components is 4 A
