# Electricity: ammeter range calculation

1. Apr 7, 2017

### Epoch

1. The problem statement, all variables and given/known data

An ammeter has a range of 10mA or 0.01A and a resistor of 20Ω.
Three shunts are used:
R1 = 4Ω ; R2 = 2Ω ; R3 = 4Ω
What will the different ranges be (m3, m2, m1)?

3. The attempt at a solution

I started by calculating the voltage:
Ua = Ia ⋅ Ra
Ua = 0.01A ⋅ 20Ω = 0.2V

Then I started calculating the current for each resistor

I1 = Ia ⋅ Ra / R1 = 0.01A ⋅ 20Ω / 4Ω = 0.05A
I2 = Ia ⋅ Ra / R2 = 0.01A ⋅ 20Ω / 2Ω = 0.1A
I3 = Ia ⋅ Ra / R3 = 0.01A ⋅ 20Ω / 4Ω = 0.05A

And here is where I get lost.
On my answer sheet it says that:
m3 = 0.075A ; m2 = 0.05A ; m1 = 0.03A

Can someone explain me what I'm missing here?
Thanks for reading up until here.

Last edited: Apr 7, 2017
2. Apr 7, 2017

### BvU

Hello Epoh,

For example: m1: if 10 mA goes through the meter, how much goes through the shunts ?

3. Apr 7, 2017

### Epoch

I don't know if I'm right but I found something:
If we take m1 that equals 0.03A and put it in the ohm's law we get:
R = Ua / m1 = 0.2 / 0.03 = 6.667Ω

If we find the equivalent resistor for serial we get:
Rt = R1 + R2 + R3 = 10Ω

Then the equivalent resistor for parallel with the serial above:
Rt = 20 ⋅ 10 / 20 + 10 = 6.667Ω

Then again the ohm's law:
m1 = Ua / Rt = 0.2 / 6.667 = 0.0299A
Then indeed it is 0.03A, but this doesn't work with the rest.
A coincidence or am I getting close to the answer?

I'm seriously stuck.

4. Apr 7, 2017

### Staff: Mentor

For a full scale reading on the ammeter its current is specified to be 10 mA, so that's a fixed quantity for each of the cases.

For each case determine which resistances are carrying only this 10 mA current (obviously the ammeter 's 20 Ω resistance is always one of them). You may want to make a sketch of the circuit and draw in the currents. Here's an example for the "m2" case. The others can be worked in a similar fashion, but this "middle" case will show the method:

Note the path of the meter current (red) and which components it flows through. Effectively, those components are in series. The remaining resistors then make up the shunt resistance for this case. It has a separate current (blue).

Does that give you any ideas? Can you work from there for this case?

5. Apr 8, 2017

### Epoch

So I tried, I don't know if I'm right,

I1 (blue) = 0.05A
Red through R2 = 0.01A
Ia = 0.01

I1 + red through R2 and R1 + Ia = 0.05 + 0.01 + 0.01 = 0.07
m3 = 0.07

R = R1 + R2 = 4 + 2 = 6Ω
I blue = U / R = 0.2 / 6 = 0.03A
Red through R3 = 0.01A
Ia = 0.01

I blue + red through R2 + Ia = 0.03 + 0.01 + 0.01 = 0.05
m2 = 0.05

R = R1 + R2 + R3 = 4 + 2 + 4 = 10Ω
I (blue) = U / R = 0.2 / 10 = 0.02A
Ia = 0.01

Ia + I (blue) = 0.01 + 0.02 = 0.03A
m1 = 0.03

It's close to my answer sheet, maybe I'm something missing.

Last edited: Apr 8, 2017
6. Apr 8, 2017

### Staff: Mentor

Okay, the problem is that you're assuming that the voltage across the shunt resistor(s) in each case is the same 0.2 V. It is not.

The voltage across the shunt is the same as the voltage that occurs across the combined meter and series resistance that forms the "red" current. Here's a modified view of the middle case, m2, where the resistance that is part of the meter's path has been moved to highlight their relationship with the shunt resistance:

Now, look at the red path where you know the given current is 10 mA: What's the potential drop from node 0 to m2?

7. Apr 8, 2017

### Epoch

I think I got it.
I calculated m2 and it seems that It's right.
Since the voltage in parallel is the same in all resistors that are in parallel I came with this answer.

I also tried m3 with the same method:

8. Apr 8, 2017

### Staff: Mentor

Excellent! Well done.

9. Apr 8, 2017

### Epoch

Sir, I thank you for your help.