Team:Grenoble-EMSE-LSU/Project/Instrumentation

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<p align="center"><img src="https://static.igem.org/mediawiki/2013/a/ab/Eq_V_circuit.PNG" alt="Voltage applied resistors" width="350px" /></br></br></p>
<p align="center"><img src="https://static.igem.org/mediawiki/2013/a/ab/Eq_V_circuit.PNG" alt="Voltage applied resistors" width="350px" /></br></br></p>
<p>The value of the resistor is also:<img src="https://static.igem.org/mediawiki/2013/9/92/Eq_R_circuit.PNG" alt="Determination_of_R" width="350px" /></br></br>
<p>The value of the resistor is also:<img src="https://static.igem.org/mediawiki/2013/9/92/Eq_R_circuit.PNG" alt="Determination_of_R" width="350px" /></br></br>
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In order for the resistor not to burn, the power dissipated due to the Joule effect need to be under 0.25W. But it is not the case here!  P=UI=RI²=1.4 x 0.5² = 0.35W > 0.25W</br>
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In order for the resistor not to burn, the power dissipated due to the Joule effect need to be under 0.25W.</br>
 +
But it is not the case here!  P=UI=RI²=1.4 x 0.5² = <strong><I>0.35W > 0.25W</I></strong></br>
To solve this issue we use a little trick. We put two resistors in parallel that have 2xR=2.8 Ω</br>
To solve this issue we use a little trick. We put two resistors in parallel that have 2xR=2.8 Ω</br>
So that we divide the power dissipated due to the Joule effect by two but keep the same intensity Ie.</br>
So that we divide the power dissipated due to the Joule effect by two but keep the same intensity Ie.</br>

Revision as of 23:50, 1 October 2013

Grenoble-EMSE-LSU, iGEM


Grenoble-EMSE-LSU, iGEM

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