Circuits question AQA unit 1 physics

Hello, I'm having a hard time working out part (c) for this question (I've attached a photo of the circuit and the question) Could someone brilliant give me hand??

Thanks :smile:

(edited 9 years ago)

Original post by Sutsugua

Hello, I'm having a hard time working out part (c) for this question (I've attached a photo of the circuit and the question) Could someone brilliant give me hand??

Thanks :smile:

Information is missing. You need to post the whole question including previous sections as this will contain the relevant information.

Reply 2

GrandMasti

11

This is a potential divider question in disguise. Each loop of the circuit is it's own potential divider - the sum of potential differences along any one loop of the circuit is equal to the sum of EMF's (which is 12V).

Reply 3

uberteknik

21

Original post by Sutsugua

Hello, I'm having a hard time working out part (c) for this question (I've attached a photo of the circuit and the question) Could someone brilliant give me hand??

Thanks :smile:

Thanks for posting the previous parts. :smile:

As the previous poster said, this is an application of potential dividers.

i.e. the voltage developed across each resistor is proportional to the ratio of the series resistances x the supply e.m.f..

In the part where it asks for the potential between the resistor pairs, you will need to find the p.d.'s referenced to the -ve terminal of the battery in both cases, then subtract one from the other.

Reply 4

Sutsugua

OP

1

Thank you both, this is very helpgul :biggrin:

However uberteknik, could you explain why you have to subtract the two p.d's in the last part of the question... I see you get the right answer but I still don't quite follow what's going on here :redface:

Thanks again :smile:

Reply 5

uberteknik

21

Original post by Sutsugua

Thank you both, this is very helpgul :biggrin:

However uberteknik, could you explain why you have to subtract the two p.d's in the last part of the question... I see you get the right answer but I still don't quite follow what's going on here :redface:

Thanks again :smile:

Voltage is defined as joules per coulomb of charge. V = energy/charge.

Potential Difference (p.d.) is therefore the difference in 'potential' to perform work between two points.

i.e. in this context, it literally means the quantitative difference between the voltages at those two points.

This can used to advantage because the thermocouple will normally be inaccessible and cannot be measured directly. Adjusting the resistance in the parallel arm of the bridge will allow the p.d. between the mid-points of both dividers to be 'nulled' (balanced). When this occurs, the ratios of the resistances in both arms must be the same.

It's then a simple case of measuring the known (reference) resistors and the actual unknown thermocouple resistance can be calculated very accurately.

Reply 6

Sutsugua

OP

1

Right, ok. So as the p.d of the 10k ohm resistor is 8 ohms and the 20k ohm resistor is 6 ohms... The p.d between C&D is 2 ohms??

Reply 7

uberteknik

21

Original post by Sutsugua

Right, ok. So as the p.d of the 10k ohm resistor is 8 ohms and the 20k ohm resistor is 6 ohms... The p.d between C&D is 2 ohms??

Referenced to the -ve terminal the p.d.'s across:

CE = 6V

DF = 4V

hence the p.d. across CD is:

(CE - DF) = (6 - 4) = 2V

Reply 8

Sutsugua

OP

1

Great! Many thanks :biggrin:

Reply 9

justalawyer1998

5

Original post by uberteknik

Referenced to the -ve terminal the p.d.'s across:

CE = 6V

DF = 4V

hence the p.d. across CD is:

(CE - DF) = (6 - 4) = 2V

Could you ever get a negative result for voltage?

Reply 10

uberteknik

21

Original post by Nikhilm

Could you ever get a negative result for voltage?

Yes, of course.

Voltage measurements are always made with respect to some reference and the +ve or -ve potential voltage is then defined by the direction of the conventional current flow:

With a battery, the reference is normally taken to be the -ve terminal, in which case the +ve terminal is stated as +volts higher potential because conventional current flows from the +ve to -ve terminals. i.e. work is done by current flowing from the higher potential to the lower potential which means the +ve terminal has a higher work potential than the -ve terminal.

(NB make sure you understand the difference between conventional current flow and electron flow directions which are opposite each other).

If the reference terminal is taken to be the +ve terminal, then the -ve terminal will be stated as -volts since conventional current flows from the +ve to the -ve terminal, hence the -ve terminal must be at a lower potential.

So it is with the bridge circuit. The potentials are stated with respect to a reference terminal and the sign of the potential is determined by the direction of the conventional current flow which could indeed be in both directions.

(edited 8 years ago)

Reply 11

justalawyer1998

5

Original post by uberteknik

Yes, of course.

Voltage measurements are always made with respect to some reference and the +ve or -ve potential voltage is then defined by the direction of the conventional current flow:

With a battery, the reference is normally taken to be the -ve terminal, in which case the +ve terminal is stated as +volts higher potential because conventional current flows from the +ve to -ve terminals. i.e. work is done by current flowing from the higher potential to the lower potential which means the +ve terminal has a higher work potential than the -ve terminal.

(NB make sure you understand the difference between conventional current flow and electron flow directions which are opposite each other).

If the reference terminal is taken to be the +ve terminal, then the -ve terminal will be stated as -volts since conventional current flows from the +ve to the -ve terminal, hence the -ve terminal must be at a lower potential.

So it is with the bridge circuit. The potentials are stated with respect to a reference terminal and the sign of the potential is determined by the direction of the conventional current flow which could indeed be in both directions.

Thanks for clearing that up! I get that now :smile:

But with the example above, I get CE = 6V and DF = 4V, and so CD = 6-4 = 2V, but what if they asked you for D-C? Would that just simply be 4-6 = -2V?

But when i think about it without the maths;
Wouldn't it be a voltage rise - i.e. because DF has a lower voltage than CE, and so D - C will increase in pd?

And in the same way, wouldn't C - D be a voltage drop since the pd is going from 6 V to 4 V? so -2V?

Reply 12

uberteknik

21

Original post by Nikhilm

Thanks for clearing that up! I get that now :smile:

But with the example above, I get CE = 6V and DF = 4V, and so CD = 6-4 = 2V, but what if they asked you for D-C? Would that just simply be 4-6 = -2V?

But when i think about it without the maths;
Wouldn't it be a voltage rise - i.e. because DF has a lower voltage than CE, and so D - C will increase in pd?

And in the same way, wouldn't C - D be a voltage drop since the pd is going from 6 V to 4 V? so -2V?

Yes D - C would be -2V.

As I said, potential is stated with respect to a reference point. D - C (D minus C) is the algebraic sum of the potentials w.r.t. the -ve supply terminal.

Circuits question AQA unit 1 physics

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