Electricity: shunt calculation for ammeter

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SUMMARY

The discussion focuses on calculating shunt resistors for an ammeter with a resistance of 20Ω and a maximum current of 10mA. The calculated shunt resistors are a = 0.1Ω, b = 0.9Ω, and c = 9Ω, which collectively yield an equivalent resistance of 10Ω. Participants emphasize the need to create simultaneous equations to solve for the individual resistances based on different current scenarios. The matrix method is utilized to derive the values of Ra, Rb, and Rc accurately.

PREREQUISITES
  • Understanding of Ohm's Law and basic electrical principles
  • Familiarity with series and parallel resistor configurations
  • Knowledge of simultaneous equations and matrix methods
  • Experience with ammeter functionality and shunt resistor applications
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  • Study the application of Ohm's Law in circuit analysis
  • Learn about series and parallel resistor calculations
  • Explore methods for solving simultaneous equations, particularly matrix methods
  • Investigate the design and application of shunt resistors in ammeter circuits
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Electrical engineering students, hobbyists working with ammeter circuits, and professionals involved in circuit design and analysis will benefit from this discussion.

Epoch
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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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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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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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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.
 

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