nmjasdk
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hi ive got a few questions that ive attached one of them is split in 2, its just a few stuff thats been confusing me. thanks. the first 2 pictures are part 1 and part2 respectively of the question. the rest are individual questions.
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uberteknik
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(Original post by nmjasdk)
hi ive got a few questions that ive attached one of them is split in 2, its just a few stuff thats been confusing me. thanks. the first 2 pictures are part 1 and part2 respectively of the question. the rest are individual questions.


Drawing from terminals A to B: place the 100 ohms resistor in series with the ammeter then in series with the LED anode with the LED cathode connected to terminal B. (NB, these series components can be placed in any order between A and B, but the LED must be connected anode towards the battery +ve terminal and cathode towards the -ve terminal.)

Place the voltmeter in parallel with this series combination directly across terminals A and B.

This is the closest schematic I could find. All components are in the correct places except the variable resistor is missing which should be placed between the battery +ve terminal and the voltmeter/ammeter junction:

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uberteknik
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(Original post by nmjasdk)
hi ive got a few questions that ive attached one of them is split in 2, its just a few stuff thats been confusing me. thanks. the first 2 pictures are part 1 and part2 respectively of the question. the rest are individual questions.


The experiment requires the p.d. across the LED to be varied.

The best test circuit will allow the widest possible range of p.d. to be developed across the (LED + 100) ohms series resistor combination from 0V up to the supply maximum voltage.

Fig 4.3 achieves this by the variable resistor arrangement which produces a continuously variable test voltage output between 0V up to the 6V supply voltage. (A range of 6V).

The arrangement of fig 4.2. will produce a limited test voltage range between:

V_{min} = V_{supply}\big(\frac{100}{(100 + 10)}\big) = 5.45 \mathrom{V}

and

V_{max} = V_{supply}\big(\frac{100}{(100 + 0)}\big) = 6 \mathrm{V}

i.e. a range of 0.55V
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uberteknik
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(Original post by nmjasdk)
hi ive got a few questions that ive attached one of them is split in 2, its just a few stuff thats been confusing me. thanks. the first 2 pictures are part 1 and part2 respectively of the question. the rest are individual questions.



Seems a bit churlish. Perhaps your teacher thought they were a bit too untidy and not accurate enough?

The BS3939 (internationally recognised) symbols are:



2nd part:

It looks similar to the parallel equation for two resistors, but it's not the same so don't confuse the two.

The voltage developed across the potential divider combination is derived from ohms law:

I = \frac{V}{R}

and because it's a series combination, the current through all components must be the same:

I_{total} = \frac{V_{supply}}{R_{total}} = \frac{V_{supply}}{R1 + R_{750}}

The voltage developed across any series resistor (and in our case the 750 ohm resistor) is:

V_{R_{750}} = I_{total}R_{750}

therefore after substitution:

V_{R_{750}} = \frac{V_{supply}}{R1 + R_{750}}R_{750} which rearranges to

V_{R_{750}} = V_{supply}\big(\frac{R_{750}}{R1 + R_{750}}\big)
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nmjasdk
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(Original post by uberteknik)


The experiment requires the p.d. across the LED to be varied.

The best test circuit will allow the widest possible range of p.d. to be developed across the (LED + 100) ohms series resistor combination from 0V up to the supply maximum voltage.

Fig 4.3 achieves this by the variable resistor arrangement which produces a continuously variable test voltage output between 0V up to the 6V supply voltage. (A range of 6V).

The arrangement of fig 4.2. will produce a limited test voltage range between:

V_{min} = V_{supply}\big(\frac{100}{(100 + 10)}\big) = 5.45 \mathrom{V}

and

V_{max} = V_{supply}\big(\frac{100}{(100 + 0)}\big) = 6 \mathrm{V}

i.e. a range of 0.55V
i get why the fig 4.2 has a range of 055v but i dont understand how the fig 4.3 arrangement can give 0v, to be honest i dont even know what going on in that arrangement because i dont understand why theat variable resistor now has a arrow going onto it what does that even mean? its also called a potential divider in fig 4.3 but it alreaydy was a potential divider in fig 4.3 anyway.
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nmjasdk
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(Original post by uberteknik)



Seems a bit churlish. Perhaps your teacher thought they were a bit too untidy and not accurate enough?

The BS3939 (internationally recognised) symbols are:



2nd part:

It looks similar to the parallel equation for two resistors, but it's not the same so don't confuse the two.

The voltage developed across the potential divider combination is derived from ohms law:

I = \frac{V}{R}

and because it's a series combination, the current through all components must be the same:

I_{total} = \frac{V_{supply}}{R_{total}} = \frac{V_{supply}}{R1 + R_{750}}

The voltage developed across any series resistor (and in our case the 750 ohm resistor) is:

V_{R_{750}} = I_{total}R_{750}

therefore after substitution:

V_{R_{750}} = \frac{V_{supply}}{R1 + R_{750}}R_{750} which rearranges to

V_{R_{750}} = V_{supply}\big(\frac{R_{750}}{R1 + R_{750}}\big)
what worrying my about the symbol i drew is that was the symbol i drew on my actual exam test and i didnt get the mark so it wasnt my teacher it was an actual examiner that thought the symbol was wrong.

anyway also wanted to say thanks for the detailed answers honestly appreciate it.
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