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Samenvatting Elektrische schakelingen

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Samenvatting Fysica hoofdstuk 5 uit Quark 5.2

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H5
Uploaded on
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THEMA 5 ELEKTRISCHE SCHAKELINGEN
1.Schakelingen van weerstand
Een schakeling oplossen betekent voor elk element bepalen welke spanning er over staat en welke stroom er door
gaat.
Schakelingen binnenin dienen om lampjes etc. te voorzien van een juiste stroom en spanning.
In een schakeling kunnen de weerstanden in serie, parallel of gemend geschakeld worden.
De hoofdkring bevat de spaningsbron, in knooppunten komen minstens 3 geleiders samen.
Woordenschat: stroomkringen, elektrisch netwerk, serieschakeling, hoofdkring, parallelschakeling, knooppunten,
deeltakken en gemengde schakelingen
AC: : wisselstroom, er is geen vaste beweging van + naar -
DC: = : gelijkstroom, er is een vaste beweging van + naar –

2.Serieschakeling
In een serieschakeling is de stroomsterkte I in elk punt gelijk. (I= I2= I3…) = eerste wet
In een serieschakeling is de bronspanning verdeeld in deelspanningen U Ri over de weerstanden Ri  Ub= UR1 + UR2 +
… = tweede wet
Wij verbruiken energie, maar geen stroom!!
Ui
 hoe groter R, hoe groter I  U i R i→ =Cte
Ri
 Bewijs 1
1.1. substiutieiirsind and eiirsind id d sir i
een equivalente schakeling die bestaat uit 1 weerstand = substtuteweerstand
 de substtuteweerstand Rs van weerstanden R1, R2… Rn in serie is gelijk aan de som van de weerstanden:
Rs= R1+ R2 + … + Rn
1.2. ioipnss dg: iid spndd dgs ilir
p101-102

3.Parallelschakeling
De stroom I wordt verdeeld over de weerstanden  I = I1 + I2 + I3… = eerste wet
De spaning U over elke tak hetzelfde  Ub= UR1 = UR2 = … = U = tweede wet
 hoe groter R, hoe kleiner Rs  Rs is altjd kleiner dan de kleinste R
 Bewijs 2
3.1. substiutieiirsind and pnrnllilli eiirsind id
de substtuteweerstand Rs van weerstanden R 1, R2 … Rn in parallel bereken je met volgende formule:
1 1 1 1
= + + …+
R s R 1 R2 Rn
3.2. ioipnss dg: iid siroom ilir
p105
R60,44
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