Electricity: shunt calculation for ammeter

In summary, the different ranges are 3000mA, 300mA, 30mA.The different resistors are a, b and c.The equivalent resistance is 10Ω.I hope someone can help me with this.
  • #1
Epoch
24
0

Homework Statement


We have an ammeter with 20Ω resistance and a range of 10mA or 0.01A.
The different ranges are 3000mA, 300mA, 30mA.
Calculate the shunt resistors a, b and c.
Problem.jpg


Homework Equations


voltage.png

shunt formula.png

shunt calculation.png

formula.png


The Attempt at a Solution


Ia = 0.01A ; Ra = 20Ω ; I = 0.03A

Ua = Ia ⋅ Ra = 0.01A ⋅ 20Ω = 0.2V
Is = I - Ia = 0.03A - 0.01A = 0.02A
current flow.jpg


The multiplication factor:
m = I / Ia = 0.03 / 0.01 = 3

I found the equivalent resistance for all shunts:
Rn = Ra / m - 1
Rn = 20 / 3 - 1 = 10Ω

In parallel the voltage is the same in all resistors, but a, b and c are in series and have their own voltages out of our Ua = 0.2V.
How can I find the voltage for a, b and c?
Otherwise I can't calculate each resistor.

The answers on my answer sheet are:
a = 0.1Ω
b = 0.9Ω
c = 9Ω
So indeed the equivalent resistor is 10Ω.

I hope someone can help me with this.
 
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  • #2
Hello again,
Epoch said:
In parallel the voltage is the same in all resistors, but a, b and c are in series and have their own voltages out of our Ua = 0.2V.
How can I find the voltage for a, b and c?
Case m1 is easy, right ? Can peek at the previous thread to help you find Ra+Rb+Rc .

Case m2: 10 mA through Rc + 20 ##\Omega##, 290 mA through the parallel path of (Ra+Rb). Gets you a ratio.
Case m3 gets you a ratio too. Two ratios and one sum: means you can solve for the three unknown Rx
 
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  • #3
You'll need to create a set of simultaneous equations to solve. For example, for the first case you found that the sum of all the shunt resistors is 10Ω : ##R_a + R_b + R_c = 10##. You need to find similar equations for the other two cases and solve for the individual resistance values.Edit. Ha! I see that BvU beat me to it!
 
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Likes Epoch
  • #4
BvU said:
Hello again,
Case m1 is easy, right ? Can peek at the previous thread to help you find Ra+Rb+Rc .

Case m2: 10 mA through Rc + 20 ##\Omega##, 290 mA through the parallel path of (Ra+Rb). Gets you a ratio.
Case m3 gets you a ratio too. Two ratios and one sum: means you can solve for the three unknown Rx
gneill said:
You'll need to create a set of simultaneous equations to solve. For example, for the first case you found that the sum of all the shunt resistors is 10Ω : ##R_a + R_b + R_c = 10##. You need to find similar equations for the other two cases and solve for the individual resistance values.Edit. Ha! I see that BvU beat me to it!

Okay, so I came up with this:
Ra + Rb + Rc = 10
30Rc = 270
300Rb +300Rc = 2970

I used the matrix method:

A
1 1 1 Ra 10
0 0 30 ⋅ Rb = 270
0 300 300 Rc 2970

Determinant of A = -9000
-----------------------
D-Ra
10 1 1
270 0 30
2970 300 300

Determinant of D-Ra = -900
--------------------------
D-Rb
1 10 1
0 270 30
0 2970 300

Determinant of D-Rb = -8100
----------------------------------------
D-Rc
1 1 10
0 0 270
0 300 2970

Determinant of D-Rc = -81000
------------------------------------------

Ra = D-Ra / A = -900 / -9000 = 0.1Ω
Rb = D-Rb / A = -8100 / -9000 = 0.9Ω
Rc = D-Rc / A = -81000 / -9000 = 9Ω

So I think I got it right.
Thanks for your time guys.
 

What is electricity?

Electricity is a form of energy that is created by the movement of electrons. It can be harnessed and used to power various devices and machines.

What is a shunt calculation for ammeter?

A shunt calculation for ammeter is a method for measuring current by diverting a known fraction of the current through a low-resistance shunt and using Ohm's law to determine the total current in a circuit.

How do I calculate the shunt resistance for an ammeter?

To calculate the shunt resistance for an ammeter, you need to know the full-scale current rating of the ammeter and the desired shunt resistance value. Then, you can use the formula Rshunt = (Ifullscale x Rload) / (Imeasured - Ifullscale), where Rload is the resistance of the load in the circuit, Imeasured is the actual current being measured, and Ifullscale is the full-scale current rating of the ammeter.

Why is shunt calculation important for ammeters?

Shunt calculation is important for ammeters because it allows for accurate measurement of current in a circuit. By using a shunt, only a small portion of the current is diverted, reducing the risk of damage to the ammeter and ensuring that the measured current is not significantly affected by the presence of the ammeter in the circuit.

Can shunt calculation be used for other types of meters?

Yes, shunt calculation can be used for other types of meters, such as voltmeters and wattmeters. It is a common method for measuring current in circuits where the full-scale current is too high for traditional meter designs. However, the specific calculations and shunt resistance values may differ depending on the type of meter and the desired measurement.

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