Ammeter between 2 branches parallel resistors

I 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.

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.

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?

I'm trying to find the reading on the ammeter in terms of R and V

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?

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.

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.

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.

I 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