Exercicios Revisao 3
April 5, 2023 | Author: Anonymous | Category: N/A
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3 : 1) 1 BC 10, 50%, =100, =50, =1Ω, =4 =40. C:
) ) ) C
) ) ) ) ) )
C C , , ,
: )
T =
1
f
= 100 µ s
D = 0 ,5 V o )
)
= D ⋅ E = 0 ,5 ⋅ 100 = 50V
V DRL( pico ) = E = 100V V o
= Ec + R ⋅ I o
I o
=
V o − Ec R
=
50 − 40 1
= 10 A
)
τ =
L R
=
4m 1
= 4m
tc = 50 µ s ta = T − tc = 100 µ − 50 µ = 50 µ s
i E (t ) = I M I M
− tc
= I m ⋅ e
τ
+
( E − Ec ) R
− 50u = I ⋅ e 4m ⋅ 1 − e τ m − tc
= I m ⋅ 987 ,58m + 745 ,33m
− ta
τ − m = I M ⋅ e
Ec
− 50u = I ⋅ e 4 m ⋅ 1 − e τ M R − ta
I m= I M ⋅ 987 ,58m − 496 ,89m I m= 9 ,69 A I M
= 10 ,31 A
)
I E = I S
= D ⋅ I o = 0 ,5 ⋅ 10 = 5 A
P E = I E ⋅ E = 5 ⋅ 100 = 500W )
Pr = (V o − Ec ) ⋅ I o
)
P Ec
)
)
+
(100 − 40)
I D
= 10 ⋅ 10 = 100W
= Ec ⋅ I o = = 400W 40 ⋅ 10
= (1 − D ) ⋅ I o = 0 ,5 ⋅ 10 = 5 A
−
40 1
− 50u 4m ⋅ 1− e
1
1− e
− 50u
4m
2) B 2, =100. A 1% . =0,75, : ) ) ) )
A Δ A A
) A ) A ) =100, =1, =200Ω
:
)
Eo
=
E )
)
)
)
∆ I =
⇒ Eo =
1 − D
E Lf
D =
I Dmed = Io = Po = Eo R I Lmed =
)
)
1
I M
=
I m
=
2
1 − D
Eo
400
R
200
=
200
I o
I o
(1 − D )
C = D ⋅ Io f ⋅ ∆vo
+
−
=
1 − 0 ,75
= 400V
= 2 A
2
(1 − D) (1 − 0 ,75)
(1 − D )
100
= 800W
=
I o
=
100 ⋅ 0 ,75 = 0 ,75 A 1m ⋅100k
400 2
=
E
D ⋅ E 2 ⋅ L ⋅ f
= 8 A
= 8+
0 ,75 ⋅ 100
2 ⋅ 1m ⋅ 100k
= 8 ,375 A
D ⋅ E
= 8 − 0 ,375 = 7 ,63 A 2 ⋅ L ⋅ f
D ⋅ Io f ⋅ ∆vc
=
0 ,75 ⋅ 2 100k ⋅ 4
= 3 ,75uF
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