cuadernillo electronica

EJERCICIOS

1. Convertir de V a µV, nV, MV

a) 30 V
30 V ×1000000 = 30000000 µV
30 V × 1000000000 = 30000000000 nV
30 V / 1000000 = 0.00003 MV

b) 5889.59 V
5889.59 V × 1000000 = 5889590000 µV
5889.59 V × 1000000000 = 5889590000000 nV
5889.59 V / 1000000 = 0.00588959 MV

c) 12 V
12 V × 1000000 = 12000000 µV
12 V × 1000000000 = 12000000000 nV
12 V / 1000000 = 0.000012 MV

2.Convertir de KV a V, MV, NV

a) 13.50 KV
13.50 KV × 1000 = 13500 V
13.50 KV / 1000000 = 0.0135 MV
13.50 KV / 1000000000 = 0.0000135 nV

b)3.7 KV
3.7 KV × 1000 = 3700 V
3.7 KV / 1000000 = 0.0037 MV
3.7 KV / 1000000000 = 0.0000037 nV

c) 45 KV
45 KV × 1000 = 45000 V
45 KV / 1000000 = 0.045 MV
45 KV / 1000000000 = 0.000045 nV

3. Convertir de Ω a KΩ

a) 389 Ω = 389 Ω / 1000 =0,389 KΩ
b) 58963214 Ω = 58963214 Ω / 1000 = 589863,214 KΩ
c) 4,69 Ω = 4,69 Ω / 1000 = 0,00469 KΩ

4. Convertir de mΩ a MΩ

a) 4892,84 mΩ
4892.84 mΩ / 1000000 = 0.00000489284 MΩ
b) 789,478 mΩ
789.478 mΩ / 1000000 = 0.000000789478 MΩ
c) 78,89 mΩ
78.89 mΩ / 1000000 = 0.00000007889 MΩ
5. Convertir de nA a A

a) 9878,789 nA
9878.789 nA / 1000000000 = 0.000000009878789 A
b) 4785 nA
4785nA / 1000000000 = 0.000004785 A
c) 478 nA
478 nA / 1000000000 = 0.000000478 A

6. Convertir de mA a MA

a) 7,89 mA
7.89 mA / 1000000 = 0.00000000789 A
b) 7845454,787 mA
7845454.787 mA / 1000000 = 0.007845454787 A
c) 54,78 mA
54.78 mA / 1000000 = 0.00000005478 A


POR MEDIO DEL CÓDIGO DE COLORES DAR VALOR A LAS SIGUIENTES RESISTENCIAS E INDICAR SU RANGO DE UTILIDAD.

a. Rojo,naranja, negro, plata= 23 Ω / 10% ; (+25.3Ω) (-20.7Ω)(23×10%=2.3, 23+2.3=25.3Ω, 23-2.3=20.7Ω)

b. Violeta, verde, café= 750Ω / 20% ; (+900Ω) (-610Ω) (750×20%= 150, 750+150= 900Ω, 760-150= 610Ω)

c. Gris, blanco, rojo, dorado= 8900Ω / 5% ; (+9345Ω) (-8455Ω) (8900×5%= 445 ; 8900+445=9345Ω, 8900-445= 8455Ω)

d. Verde, amarillo, naranja, plata= 54000Ω / 10% ; (+59400Ω) (-48600Ω) (54000×10%= 5400,54000+5400= 59400Ω, 54000-5400= 48600Ω)

e. Café, rojo, café= 120Ω / 20% ; (+132Ω) (-108Ω) (120×10%= 12, 120+12= 132Ω, 120-12= 108Ω)

f. Negro, azul, naranja, dorado= 6000Ω / 5% ; (+6300Ω) (-5700Ω) (6000×5%= 300, 6000+300= 6300Ω, 6000-300= 5700Ω)

g. Blanco, amarillo, negro= 94Ω /20% ; (+112.8Ω) (-75.2Ω) (94×20%= 18.8, 94+18.8= 112.8Ω, 94-18.8= 75.2Ω)

h. Azul, azul, café, plata= 660Ω/ 10% ; (+726Ω) (-594Ω) (660×10%= 66, 660+66= 726Ω, 660-665= 594Ω)

i. Café, café, rojo, dorado= 1100Ω / 5% ; (+1155Ω) (-1045Ω) (1100×5%= 55, 1100+55= 1155Ω, 1100-55= 1045Ω)

j. Verde, violeta, café= 570Ω / 20% ; (+684Ω) (-456Ω) (570×20%= 114, 570+114= 684Ω, 570-114= 456Ω)


RESOLVER LAS SIGUIENTES INCOGNITAS

V=?; I = 2A, R= 8Ω, (V= I×R= 2A×8Ω= V16)
I= 5A, 8V, R=?,(R= V/I= 8V/5A= R1.6 Ω)
I=?; V=110V, R=20Ω, (I =V/R= 110V/20 Ω= 5.5A)
P=?; V=80V, R=25 Ω, (P=V2/R= V 802/25 Ω= V 6400/25 Ω=256W)
R=?; V=440V, I= 120A, (R=V/I=V 440/ 120A= 3.7 Ω)
V=?; P=40W, I=80A, (V=P/I= 40W/80A= 0.5V)
R=?; P=40W, I=9A, (R=P/I2= 40W/ 92A= 40W/81A=0.50 Ω)


RESOLVER LOS SIGUIENTES CIRCUITOS EN SERIE DE MANERA COMPLETA Y REALIZAR EL CIRCUITO

(ENCONTRAR RT, IT, VT, PT)

I.V= 110 V, R1= 20Ω, R2= 10Ω, R3= 30Ω

RT=R1+R2+R3= 20Ω+10Ω+30Ω= 60Ω
IT= VT/RT= 110V/60Ω= 1.83A
V1=IT×R1= 1.83×20Ω=36.6V
V2=IT×R2= 1.83×10Ω=18.3V
V3=IT×R3= 1.83×30Ω=54.9V
VT=V1+V2+V3= 36.6V+18.3V+54.9V=109.8V
P1= V1×IT= 36.6V×1.83A= 66.97W
P2= V2×IT= 18.3V×1.83A= 33.48W
P3= V3×IT= 54.9V×1.83A= 100.46W
PT=VT×IT= 109.8V×1.83A=200.9W
PT= P1+P2+P3= 66.97W+33.48W+100.46W= 200.9W


II. V=80 V, R1=12Ω, R2=22Ω, R3=32Ω, R4=42Ω

RT=R1+R2+R3+R4= 12Ω+22Ω+32Ω+42Ω=108Ω
IT= VT/RT=80V/108Ω=0.74A
V1=IT×R1= 0.74A×12 Ω=8.8V
V2=IT×R2= 0.74A×22 Ω=16.28V
V3=IT×R3= 0.74A×32 Ω=23.68V
V4=IT×R4=0.74A×42 Ω=31.08V
VT=V1+V2+V3+V4= 8.8V+16.28V+23.68V+31.08V=79.84V
P1= V1×IT= 8.8V×0.74A= 6.51W
P2= V2×IT= 16.28V×0.74A= 12.04W
P3= V3×IT= 23.68V×0.74A= 17.52W
P4= V4×IT= 31.08V×0.74A= 22.99W...