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everything maths and science - C2B2A

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Stap 2: Bepaal hoe om die probleem te benader<br />

Aangesien die resistors ohmies van aard is, kan ons Ohm se wet gebruik. Daar is egter<br />

twee resistors in die stroombaan; ons sal die totale weerst<strong>and</strong> moet bepaal.<br />

Stap 3: Vind die ekwivalent weerst<strong>and</strong> vir die stroombaan<br />

Omdat die resistors in parallel gekoppel is, is die totale (ekwivalente) weerst<strong>and</strong> R is<br />

Stap 4: Pas Ohm se wet toe<br />

Stap 5: Skryf die finale antwoord<br />

1 1<br />

= +<br />

R R1<br />

1<br />

.<br />

R2<br />

1 1<br />

= +<br />

R R1<br />

1<br />

R2<br />

= 1 1<br />

+<br />

2 4<br />

= 2+1<br />

4<br />

= 3<br />

4<br />

Daarom, R = 1,33Ω<br />

R = V<br />

I<br />

R · I V<br />

=<br />

R I<br />

· I<br />

R<br />

I = V<br />

R<br />

I = V · 1<br />

R<br />

<br />

3<br />

= (12)<br />

4<br />

=9A<br />

Die stroom wat in die stroombaan vloei is 9 A.<br />

Uitgewerkte voorbeeld 6: Ohm se wet, parallele stroombaan<br />

VRAAG<br />

394 11.2. Ohm se wet

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