How do I measure the "load" in a schematic?

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The discussion focuses on understanding the concept of "load" in a circuit schematic, particularly in relation to EMF and internal resistance. Load refers to components that draw current, excluding internal resistance, and is not a fixed value. Participants clarify that terminal voltage is the voltage across the load, while EMF represents the battery's voltage without current flow. There is some confusion regarding the placement of the voltmeter, with the consensus that it should be placed across the load to measure voltage accurately. The internal resistance of the battery will influence any voltmeter readings, regardless of its placement.
Tasdel

Homework Statement


Schematic..PNG

Im doing a project on Emf and internal resistance. I'm just trying to understand what the load would be and where I would put the positive and negative pens on the schematic to find the load.

Homework Equations


Terminal Voltage would be what I have in the image above correct? the 8.24V.
and the EMF is the Voltage of the battery without current passing through which would be 9V. (is that right?)
The load is any resistance as well not including the internal resistance?

The Attempt at a Solution


My tiny bit of understanding on the load is its anything that draws current or uses energy.
This would be my guess at the load.
Schematic 2.PNG
 

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Tasdel said:
what the load would be
The way I understand the term, load refers to one or more components of the circuit. It is not something which has a value. You might want the resistance of a load, or the power it is to consume.

The first image above seems to be overspecified. You have all the resistances in the circuit, including internal resistance, and the nominal voltage. From those you should be able to compute the actual voltage across the terminals, but you are given that as well. So it is not clear to me what you are trying to work out.

I have no idea how your second image constitutes "a guess at the load".
 
haruspex said:
The way I understand the term, load refers to one or more components of the circuit. It is not something which has a value. You might want the resistance of a load, or the power it is to consume.

The first image above seems to be overspecified. You have all the resistances in the circuit, including internal resistance, and the nominal voltage. From those you should be able to compute the actual voltage across the terminals, but you are given that as well. So it is not clear to me what you are trying to work out.

I have no idea how your second image constitutes "a guess at the load".
haruspex said:
The way I understand the term, load refers to one or more components of the circuit. It is not something which has a value. You might want the resistance of a load, or the power it is to consume.

The first image above seems to be overspecified. You have all the resistances in the circuit, including internal resistance, and the nominal voltage. From those you should be able to compute the actual voltage across the terminals, but you are given that as well. So it is not clear to me what you are trying to work out.

I have no idea how your second image constitutes "a guess at the load".
The load is is every resistor outside the battery and that means also not the internal resistance. I was thinking because the voltmeter was measuring right next to the negative end of the battery that it was also taking into account the resistance of the battery.
 
The main question was just asking me to place the voltmeter across the load. So would that be the first image I posted? I am assuming it could be because its getting the whole circuit
 
Tasdel said:
The main question was just asking me to place the voltmeter across the load. So would that be the first image I posted? I am assuming it could be because its getting the whole circuit
Yes.
 
Tasdel said:
because the voltmeter was measuring right next to the negative end of the battery that it was also taking into account the resistance of the battery.
No matter where you put the voltmeter (that gives a nonzero reading) the reading will be influenced by the internal resistance. Indeed, the internal resistance would be computable from it.
 
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