Saber Hacer: 1. ¿Cuál es la capacitancia de un capacitor que almacena 12.35 C a 3V?
C=Q/V C= 12.35C / 3V = 4,1167F
2. Encuentre las cantidades indicadas:
Capacitancia (F) | Carga (Q) | Voltaje (v) | 0,1 | 11 | 110 | 0.3 | 66 | 220 | 0.2 | 50 | 0.004 |
C=Q/V = 11C/110V = 0.1F Q=V.C = 220V.0.3F = 66C V=C/Q = 0.2F/50C = 0.004V
3. Hallar la capacitancia total para:
[pic 3] CT= C1+C2+C3 CT= 0.12µF + 0.008µF + 0.5mF CT= 0.0005002F = 5002mF
| [pic 4] CT= 1 [pic 5][pic 6] 1/0.12µF +1/0.008µF+1/0.5Mf CT= 8,333 + 12,500 + 2,000 = 22,833 CT= 1/ 22,833 = 43,7963µF CT= 43,7963µF
| [pic 7] CT= C2+C3 = 0,08µF + 0,5mF =0,50008mF
[pic 8]
CT= 1/0.12µF+ 1/0,50008mF+1/150pF CT= 8,333 + 1,999.68+6666666 CT= 1/6666676,33268 CT= 1,49999F
| [pic 9] CT=1/0,12 µF+1/0,08µF+1/0,5mF+1/150pF + 1/0,075µF
CT= 8,333+12,500+2000+6666666+13333333 CT= 1/ 20002019,833 CT= 4,9994950F
| [pic 10] CT= 1/ 0,08µF + 1/0,5mF+1/150pF CT= 12,500+2000+666666 CT= 668678,5F
CT= 215pF+ 668678,5F = 668678F CT= 1/0,12µF+1/668678F+1/0,075µF CT=8,333+1,4954+13333333 CT= 1/13333342,8284 CT= 7,4999946F
| C1=0.12 µF C2=0.08 µF C3= 0.5 mF C4= 150 pF C5= 0.075µF C6= 215 pF |
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