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Bron in tak 1 kortsluiten, maar Ri laten staan<br />
I1“ = 750 mA<br />
I2“ = 31,25 mA<br />
I3“ = 781,25 mA<br />
Maak tenslotte de algebraïsche som van alle deelstromen<br />
De grootste deelstroom bepaalt de uiteindelijke zin.<br />
<br />
I1 = I1’ – I1” = 800 – 750 = 50 mA <br />
<br />
I2 = I2’ + I2” = 50 + 31,25 = 81,25 mA <br />
<br />
I3 = I3” – I3’ = 781,25 – 750 = 31,25 mA <br />
KHLim – Lerarenopleiding – Superpositiemethode TR 2
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