Electrical Circuit - Power dissipation

AI Thread Summary
In the lab, the purpose of measuring the source voltage with the circuit disconnected is to obtain the unloaded voltage output of the power supply, which reflects the ideal voltage without any current draw. When the circuit is connected, the internal resistance of a non-ideal DC power source causes voltage drops, leading to inaccurate power dissipation readings. Understanding the difference between ideal and non-ideal sources is crucial, as the latter includes internal losses that affect voltage output. The discussion highlights that measuring voltage while connected would also account for power dissipated by the power supply itself, complicating the results. Accurate power dissipation measurements require isolating the circuit to avoid these additional losses.
rambo5330
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During a Lab we had a simple purely resistive circuit hooked up to a 10 VDC power supply with an adjustable voltage output. In order to determine the power dissipated by the circuit we were asked to determine the source voltage while the circuit was DISCONNECTED... we were not told why... In the Lab write up we are asked to explain clearly why we had to do this...

The only thing I can think of is that with the circuit disconnected no current would be running through the secondary side of the step down transformer in the power supply which would give you the unloaded voltage output...when the circuit is connected current would start flowing throught he secondary and maybe the CEMF affects the apparent voltage output? I have no idea... Can someone please explain why this gives a more accurate reading for power dissipated?
 
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What is the essential difference between an ideal DC voltage source and a non-ideal DC voltage source?
 
im not 100% sure what you mean by that but I am assuming you mean... an ideal DC power source has 0 losses and a non-ideal power source does have losses?
 
rambo5330 said:
im not 100% sure what you mean by that but I am assuming you mean... an ideal DC power source has 0 losses and a non-ideal power source does have losses?

Essentially, yes. Basically, any non-ideal DC power source will operate similarly to an ideal DC power source except that it has an internal resistance. How might you apply that knowledge to the problem at hand?
 
actually that is exactly what your talking about... i just drew it out with the source and a resistor in series and did KVL and yes i totally understand now... but basically this is not giving true power dissipated by the circuit is it not... its now including power dissipated by the power supply as well... (which isss part of the circuit i suppose).. could we not just adjust the voltage to our probler 10 V while the circuit is attached and then measure the current at various parts and use this as the power dissipated by our circuit ...
 
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