Ammeter between 2 branches parallel resistors

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alvan15
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Size:  451.4 KBI have applied kirchoff current law , also the p.d between the ammeter should be 0 as it has 0resistance, from there I have no idea what to do
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uberteknik
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(Original post by alvan15)
Name:  1507054782986617537995.jpg
Views: 679
Size:  451.4 KBI have applied kirchoff current law , also the p.d between the ammeter should be 0 as it has 0resistance, from there I have no idea what to do
Hello.

I don't know what problem you are trying to solve?

Notice that there are two potential divider pairs of resistors, with each pair in parallel placed across the battery terminals.

The p.d. across the ammeter can only be 0V when the p.d. at the centre of each divider pair is exactly the same with reference to a common node where the pairs are connected.
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alvan15
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(Original post by uberteknik)
Hello.

I don't know what problem you are trying to solve?

Notice that there are two potential divider pairs of resistors, with each pair in parallel placed across the battery terminals.

The p.d. across the ammeter can only be 0V when the p.d. at the centre of each divider pair is exactly the same with reference to a common node where the pairs are connected.

For that condition to be a achieved, the ratio of the resistances in each divider pair, must be identical.

For example, both pairs comprise resistors all of the same value R.
Or, both pairs of divider resistors are in the same ratio: R and a multiple of R say.
I'm trying to find the reading on the ammeter in terms of R and V
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alvan15
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(Original post by uberteknik)
Hello.

I don't know what problem you are trying to solve?

Notice that there are two potential divider pairs of resistors, with each pair in parallel placed across the battery terminals.

The p.d. across the ammeter can only be 0V when the p.d. at the centre of each divider pair is exactly the same with reference to a common node where the pairs are connected.

For that condition to be a achieved, the ratio of the resistances in each divider pair, must be identical.

For example, both pairs comprise resistors all of the same value R.
Or, both pairs of divider resistors are in the same ratio: R and a multiple of R say.
If i knew it is a ideal ammeter wouldn't it have 0 resistance and therefore 0p.d?
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uberteknik
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(Original post by alvan15)
If i knew it is a ideal ammeter wouldn't it have 0 resistance and therefore 0p.d?
Ahhhhhh. You are of course correct in your assumption - in an ideal world.

In reality, all real-world components are not ideal. Ammeters do have resistance, albeit, very small.

Examiners use 'ideal' as a concept to simplify the calculation and solution. In the context of this question, ideal is meant to imply 'ammeter resistance' has no effect on the circuit and any current reading made is true. The ideal ammeter does not drop any volts to function or alter the circuit and resistance can therefore be ignored as if it were zero.

In contrast, an ideal voltmeter is deemed to have infinite resistance to again simplify calculations. That is, the voltmeter does not require current to function or alter the circuit and can be ignored as if it were infinite resistance.
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uberteknik
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(Original post by alvan15)
I'm trying to find the reading on the ammeter in terms of R and V
..
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alvan15
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(Original post by uberteknik)
Find an expression for each potential divider pair individually. Use the negative terminal of the battery as a common reverence and find the p.d. across the resistors connected to that common.

Then subtract the two values to find the p.d. effectively across the ammeter and simplify.

Hint: V/2 - V/3
I know p.d is V/6 but since R=V/I , wouldn't that mean the reading would be 0A since R is 0?
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uberteknik
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(Original post by alvan15)
I know p.d is V/6 but since R=V/I , wouldn't that mean the reading would be 0A since R is 0?
Hi,

Having read my previous answer again, I need to make a correction. Apologies for this, I guess it was too early in the morning and no breakfast!

You are of course absolutely correct in your assertion that an ammeter resistance of zero ohms, would mean no potential difference can be developed across it.

In which case, the nodes of the potential dividers must also be at the same potential by virtue of the ammeter creating a short circuit between them.

This means the circuit comprises two sets of parallel resistors connected in series:

The first pair of R in parallel with 2R ...R_{left} = \frac{2R \mathrm x R}{2R \mathrm + R} = \frac{2R^{2}}{3R} = \frac{2R}{3}

With p.d. across them of ...V_{left} = V(\frac{\frac{2R}{3}}{\frac{2R}{3}+\frac{R}{2}}) = \frac{4V}{7}

This is in series with,

the second pair of R in parallel with R ...R_{right} = \frac{R}{2}

with a p.d. across them of \frac{3V}{7}



The p.d. across the ammeter is 0V.
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alvan15
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(Original post by uberteknik)
Hi,

Having read my previous answer again, I need to make a correction. Apologies for this, I guess it was too early in the morning and no breakfast!

You are of course absolutely correct in your assertion that an ammeter resistance of zero ohms, would mean no potential difference can be developed across it.

In which case, the nodes of the potential dividers must also be at the same potential by virtue of the ammeter creating a short circuit between them.

This means the circuit comprises two sets of parallel resistors connected in series:

The first pair of R in parallel with 2R ...R_{left} = \frac{2R \mathrm x R}{2R \mathrm + R} = \frac{2R^{2}}{3R} = \frac{2R}{3}

With p.d. across them of ...V_{left} = V(\frac{\frac{2R}{3}}{\frac{2R}{3}+\frac{R}{2}}) = \frac{4V}{7}

This is in series with,

the second pair of R in parallel with R ...R_{right} = \frac{R}{2}

with a p.d. across them of \frac{3V}{7}



The p.d. across the ammeter is 0V.
thank you so much for the help
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Mista Dobalina
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(Original post by uberteknik)
Hi,

Having read my previous answer again, I need to make a correction. Apologies for this, I guess it was too early in the morning and no breakfast!

You are of course absolutely correct in your assertion that an ammeter resistance of zero ohms, would mean no potential difference can be developed across it.

In which case, the nodes of the potential dividers must also be at the same potential by virtue of the ammeter creating a short circuit between them.

This means the circuit comprises two sets of parallel resistors connected in series:

The first pair of R in parallel with 2R ...R_{left} = \frac{2R \mathrm x R}{2R \mathrm + R} = \frac{2R^{2}}{3R} = \frac{2R}{3}

With p.d. across them of ...V_{left} = V(\frac{\frac{2R}{3}}{\frac{2R}{3}+\frac{R}{2}}) = \frac{4V}{7}

This is in series with,

the second pair of R in parallel with R ...R_{right} = \frac{R}{2}

with a p.d. across them of \frac{3V}{7}



The p.d. across the ammeter is 0V.
Thank you, this really helped me too.
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AJ113817
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(Original post by uberteknik)
Hi,

Having read my previous answer again, I need to make a correction. Apologies for this, I guess it was too early in the morning and no breakfast!

You are of course absolutely correct in your assertion that an ammeter resistance of zero ohms, would mean no potential difference can be developed across it.

In which case, the nodes of the potential dividers must also be at the same potential by virtue of the ammeter creating a short circuit between them.

This means the circuit comprises two sets of parallel resistors connected in series:

The first pair of R in parallel with 2R ...R_{left} = \frac{2R \mathrm x R}{2R \mathrm + R} = \frac{2R^{2}}{3R} = \frac{2R}{3}

With p.d. across them of ...V_{left} = V(\frac{\frac{2R}{3}}{\frac{2R}{3}+\frac{R}{2}}) = \frac{4V}{7}

This is in series with,

the second pair of R in parallel with R ...R_{right} = \frac{R}{2}

with a p.d. across them of \frac{3V}{7}



The p.d. across the ammeter is 0V.
What is the end ammeter reading in terms of R and V?
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RogerOxon
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(Original post by alvan15)
Name:  1507054782986617537995.jpg
Views: 679
Size:  451.4 KBI have applied kirchoff current law , also the p.d between the ammeter should be 0 as it has 0resistance, from there I have no idea what to do
The ammeter has zero resistance, so will pass as much current as is required to make the voltage across it zero. Note that to achieve that, the right resistors have to carry the same current.

What is the effective resistance of the circuit? Hint: The ammeter is a short.
Last edited by RogerOxon; 3 weeks ago
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