Electricity: shunt calculation for ammeter

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Homework Help Overview

The discussion revolves around calculating shunt resistors for an ammeter with a specified resistance and current range. The problem involves understanding the relationships between the shunt resistors and the ammeter's resistance in the context of electrical circuits.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the calculation of shunt resistors based on given current values and resistance. There are discussions about the voltage across each shunt resistor and the need for simultaneous equations to solve for the unknown resistances.

Discussion Status

Some participants have provided equations and methods for solving the problem, including the use of ratios and simultaneous equations. There is an ongoing exploration of how to derive the individual values of the shunt resistors based on the total resistance and current distribution.

Contextual Notes

Participants mention the need to consider different cases for the current through the shunt resistors and the total resistance, indicating a complex relationship that requires careful analysis. The original poster has provided some calculated values that are being discussed and verified.

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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