# Determining r and emf

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#1
Over here in part a) why do we have to include a resistor in the diagram? What's the point of it? I've never seen it in previous years.

If in part d) I wrote about current and ammeter what could be the appropriate scale be then? Or is there no way to gain the 2nd mark?

And if instead of pd I wrote voltage wherever the mark scheme says pd/potential difference do I lose any marks?  0
1 year ago
#2
(Original post by Presto)
Over here in part a) why do we have to include a resistor in the diagram? What's the point of it? I've never seen it in previous years.

If in part d) I wrote about current and ammeter what could be the appropriate scale be then? Or is there no way to gain the 2nd mark?

And if instead of pd I wrote voltage wherever the mark scheme says pd/potential difference do I lose any marks?
Part a) Emf = I(r + R) also known as Emf = V + Ir.
the load resistor (the resistor that's meant to be in the circuit) gives resistance R and internal resistance is r. Without a resistor in the circuit, (the cell goes from one terminal to another using one wire) then the R would be 0 meaning E = Ir, this can lead to the wires melting, a fire starting or something else. Just a safety thing I'm pretty sure, which is what you need to consider when doing experiments. Also, for the experiment you need to vary resistance in order to find the emf as well as the internal resistance. (So, I assume a variable resistor is needed.)
V= -rI + E (rearranged the formula.)
Without a resistor of any kind then V would be 0 constantly (V=IR) meaning you can't pick numerous values from the graph for the gradient and can't plot anything.

part d) I'm very unsure if you can get a current... having only the one value would mean getting current from any formula that includes voltage is impossible.

p.d and voltage are the same thing, so, yea you'd get the mark for it, I think... usually in questions you can swap them in and out with no consequences.

Spoiler:
Show

Actually these are are same thing but usage is at different places. Whenever we talk about batteries or a DC system, we use the Potential difference, as there is potential difference of 3.7 Volt. Voltage is used as Output from an electrical machine.
1
#3
(Original post by Relentas)
Part a) Emf = I(r + R) also known as Emf = V + Ir.
the load resistor (the resistor that's meant to be in the circuit) gives resistance R and internal resistance is r. Without a resistor in the circuit, (the cell goes from one terminal to another using one wire) then the R would be 0 meaning E = Ir, this can lead to the wires melting, a fire starting or something else. Just a safety thing I'm pretty sure, which is what you need to consider when doing experiments. Also, for the experiment you need to vary resistance in order to find the emf as well as the internal resistance. (So, I assume a variable resistor is needed.)
V= -rI + E (rearranged the formula.)
Without a resistor of any kind then V would be 0 constantly (V=IR) meaning you can't pick numerous values from the graph for the gradient and can't plot anything.

part d) I'm very unsure if you can get a current... having only the one value would mean getting current from any formula that includes voltage is impossible.

p.d and voltage are the same thing, so, yea you'd get the mark for it, I think... usually in questions you can swap them in and out with no consequences.

Spoiler:
Show

Actually these are are same thing but usage is at different places. Whenever we talk about batteries or a DC system, we use the Potential difference, as there is potential difference of 3.7 Volt. Voltage is used as Output from an electrical machine.

I thought the variable resistor would act as the load resistor and give the V = IR thingy. I did draw that, just not a resistor.
Thanks soo much for this! 0
1 year ago
#4
(Original post by Presto)
I thought the variable resistor would act as the load resistor and give the V = IR thingy. I did draw that, just not a resistor.
Thanks soo much for this! I think maybe a variable resistor would be fine. Its a resistor so it fits into what the question is asking for as long as there's something to stop it short circuiting I'm sure it'd be fine. Imo having a cell connected to a variable resistor with volt meter and ammeter in the right place would get you the marks.
0
#5
(Original post by Relentas)
I think maybe a variable resistor would be fine. Its a resistor so it fits into what the question is asking for as long as there's something to stop it short circuiting I'm sure it'd be fine. Imo having a cell connected to a variable resistor with volt meter and ammeter in the right place would get you the marks.
Prsom. I think you're right but this is from the latest paper so I know it won't come in my exam but just in case it does I'll add another resistor with the variable resistor just to be on the safe side. Thanks again!
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