# Measuring Range Extension "homework"

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1. Dec 5, 2017

### Apo_GER

• Moved from a technical forum, so homework template missing
a)
dimension the moving-coil movement so that it indicates full deflection at 25 V
given Values:

Voltage Source: U2Max = 25V;
moving-coil movement: IM = 100µA; UM = 270mV

Rv = ( U2Max - UM ) / IM = (25V - 270mV) / 100 µA = 247,3kΩ

b) the internal resistance is increased by 20 % due to tolerances in production. What does the moving coil movement indicate at an output voltage of 25V

My approach is the following...

RM = UM / IM = 270mV / 100µA = 2700Ω

now: it is 20% higher --> RM.Tol = RM *1,2 = 3240Ω

...........
...............

The Answer is : The moving coil movement indicates 24.945 V

2. Dec 5, 2017

### BvU

Hallo Apo,

Nor clear what your dots are representing....
You found a new value for RM and the question is: how far does the meter that you built with this coil and the 247.3 k$\Omega$ series resistance go on the scale, when the actually applied voltage is 25 V ?

(and you already know that 100 $\mu$A gives full scale)

3. Dec 5, 2017

### Apo_GER

Hi BvU
The dots are representing the missing "method" to get to the answer.

In task-part a) i came to the result, that i have to add an other resistance Rv with 247,3kOhm in series so that the moving coil movement shows full deflection.

In b) the internal resistance increases by 20% and voltage source supplies 25 V.
The question is what does the moving coil movement shows?
The answer has to be 24.945 V.

My Idea:

The inner resistance increase by 20%...
--> Rm = Um/Im --> 270mV * 100µA = 2700Ohm
--> RM.Tol = Rm * 1,2 = 3240Ohm

Now the current strength changes? or the voltage drop? "Over moving coil movement"

--> UM.Tol = RM.Tol * IM = 3240Ohm * 100µA = 324mV

--> Rv = ( U2Max.Tol - UM.Tol ) / IM = > 247,3kOhm = ( U2Max.Tol - 324mV ) / 100µA => (247,3kOhm * 100µA ) + 324mV = U2Max.Tol
=> U2Max.Tol = 25,054 (not true)

4. Dec 5, 2017

### BvU

Hehe, with the answer given you should be able to draw your own conclusion ....

The intention of he exercise is
for an applied voltge of 25 V the current is now a bit lower. The ideal coil showed full scale at 100 $\mu$A, but the coil with 20% more resistance will cause a smaller current, so it does not go all the way to full scale. With full scale (= 100 $\mu$A ) marked as 25V for the ideal coil, the non-ideal coil shows ...

extra exercise: how big is now the voltage drop over the coil ?