Give two causes that can result in surface cracking on extruded products.
1 answer:
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Given Information:
Inductance = L = 5 mH = 0.005 H
Time = t = 2 seconds
Required Information:
Current at t = 2 seconds = i(t) = ?
Energy at t = 2 seconds = W = ?
Answer:
Current at t = 2 seconds = i(t) = 735.75 A
Energy at t = 2 seconds = W = 1353.32 J
Explanation:
The voltage across an inductor is given as
The current flowing through the inductor is given by
Where L is the inductance and i(0) is the initial current in the inductor which we will assume to be zero since it is not given.
So the current at t = 2 seconds is
The energy stored in the inductor at t = 2 seconds is
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
