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Brug van Wheatstone

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Voorbeeld 4 – ½ <strong>Wheatstone</strong>brug<br />

• Gegeven: ½ brug = 2 onbekende weerstanden<br />

R1 = 6 , R3 = 8 .<br />

Totale stroom = 2 A<br />

<strong>Brug</strong> is in evenwicht<br />

• Gevraagd: R2 = R X1 = ? en R4 = R X2 = ?<br />

• Oplossing:<br />

<strong>Brug</strong> in evenwicht, men mag doorverbinden<br />

Totale weerstand:<br />

<br />

<br />

R X1 R X2<br />

U 12<br />

6 Rtot<br />

( R1 // R3<br />

) ( RX<br />

2<br />

// RX<br />

2)<br />

I 2<br />

68<br />

RX<br />

2<br />

// RX<br />

2<br />

Rtot<br />

( R1<br />

// R3<br />

) 6 2,5714<br />

6 8<br />

R1<br />

R <br />

3<br />

R1<br />

RX<br />

2<br />

6<br />

RX1 RX<br />

2<br />

0, 75<br />

RX<br />

2<br />

R<br />

R 8<br />

R tot<br />

X1<br />

R X 2<br />

R X1 en R X2 staan parallel:<br />

R X1 uit brugevenwicht berekenen:<br />

3<br />

KHLim-Lerarenopleiding Vakstudie Elektriciteit - René Peeters 26

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