# Homework Help: Electricity: shunt calculation for ammeter

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1. Apr 10, 2017

### Epoch

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

2. Relevant equations

3. 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

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.

2. Apr 10, 2017

### BvU

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

3. Apr 10, 2017

### Staff: Mentor

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!

4. Apr 10, 2017

### Epoch

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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