Engineering Finding Internal Resistance through Circuit Analysis

AI Thread Summary
The discussion focuses on determining the internal resistance of a meter movement using circuit analysis. The equation for internal resistance, R_I, is given as R_I = (R_s * R_p) / (R_s - R_p), but the derivation is unclear to the participants. They explore simplifying the circuit, particularly the parallel components, and express confusion over the total resistance and how to eliminate it from their calculations. Participants emphasize the need for clarity on the experimental values of R_s and R_p, as well as the conditions under which they were measured, to properly analyze the circuit. The conversation highlights the importance of using appropriate circuit analysis methods to find the current through the meter accurately.
Safder Aree
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Homework Statement



Given this following circuit:
upload_2018-10-21_17-29-13.png


What is the internal resistance of the meter movement ( R_I).

This is part of a project I'm doing and I know the equation that gives you the internal resistance in this circuit.
$$R_I = \frac{R_sR_p}{R_s - R_p}$$.

However, I have no idea how this is derived. I know the values for V, R_s and R_p as it was done experimentally.

Homework Equations


$$R_I = \frac{R_sR_p}{R_s - R_p}$$.

$$ V = IR $$

The Attempt at a Solution


Using circuit analysis I can simplify the parallel portion as:
$$ R_{parallel} = \frac{1}{R_p} + \frac{1}{R_I}$$

Thus:
$$R_{total} = R_s + \frac{1}{R_p} + \frac{1}{R_I} $$

But what is R_total, how do I get rid of it?

Thank you for any guidance.
 

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Safder Aree said:
Using circuit analysis I can simplify the parallel portion as:
$$ R_{parallel} = \frac{1}{R_p} + \frac{1}{R_s}$$
Why do you consider those two resistances to be in parallel? Are you trying to find a Thevenin equivalent?
Thus:
$$R_{total} = R_s + \frac{1}{R_p} + \frac{1}{R_s} $$
That can't be right, the units of the terms would not match.
But what is R_total, how do I get rid of it?
Perhaps you can explain what it is you're trying to accomplish with the steps you've taken? What's your plan of attack on this problem?

Is there more information about the lab that you can share? How did you arrive at values for ##R_s## and ##R_p##? What was the procedure?

Is there more information about the meter movement that we need to know, such as the full-scale current value?
 
gneill said:
Why do you consider those two resistances to be in parallel? Are you trying to find a Thevenin equivalent?

That can't be right, the units of the terms would not match.

Perhaps you can explain what it is you're trying to accomplish with the steps you've taken? What's your plan of attack on this problem?

Is there more information about the lab that you can share? How did you arrive at values for ##R_s## and ##R_p##? What was the procedure?

Is there more information about the meter movement that we need to know, such as the full-scale current value?

I made a typo. R_parallel is actually
$$ R_{parallel} = \frac{1}{R_p} + \frac{1}{R_I}$$

To get ##R_s## and ##R_p## I used the values from a variable resistance box. The ##R_s## value was found first in a circuit where there was no ##R_p##, the value was what gave the meter movement full scale deflection. Then I proceeded to add ##R_p## and find the resistance for when I got half scale deflection on the meter movement.
 
Safder Aree said:
To get $R_s$ and $R_p$ I used the values from a variable resistance box. The $R_s$ value was found first in a circuit where there was no $R_p$, the value was what gave the meter movement full scale deflection. Then I proceeded to add $R_p$ and find the resistance for when I got half scale deflection on the meter movement.
So you need to incorporate those conditions into your analysis. Assume some meter current ##I_m## is the full-scale current value for your meter, and that deflection is proportional to current. What equations can you write for the two scenarios?
 
gneill said:
So you need to incorporate those conditions into your analysis. Assume some meter current ##I_m## is the full-scale current value for your meter, and that deflection is proportional to current. What equations can you write for the two scenarios?

So I know that ##I_m## must be:
$$ \frac{V}{R_s + R_I}$$?

Not sure where to go from here.
 
Safder Aree said:
So I know that ##I_m## must be:
$$ \frac{V}{R_s + R_I}$$
Okay, that's for the first case where ##R_p## was not connected.

What's the situation for the next scenario where ##R_p## is introduced?
 
gneill said:
Okay, that's for the first case where ##R_p## was not connected.

What's the situation for the next scenario where ##R_p## is introduced?

Then for the case $R_p$ is introduced, it should be one half of that value right?
 
Safder Aree said:
Then for the case $R_p$ is introduced, it should be one half of that value right?
That's what your additional information about the problem stated, so yes. Can you analyze the circuit to find the current through the meter for this scenario?
 
gneill said:
That's what your additional information about the problem stated, so yes. Can you analyze the circuit to find the current through the meter for this scenario?

Then current is,

$$I = VR_{total}$$
$$= V (R_s + \frac{1}{R_p} + \frac{1}{R_I})$$
 
  • #10
Safder Aree said:
Then current is,

$$I = VR_{total}$$
$$= V (R_s + \frac{1}{R_p} + \frac{1}{R_I})$$

Sorry, That makes no sense. The units do not work in that expression; You cannot sum Ohms with inverse Ohms.

Use whatever circuit analysis method you prefer to determine the current though the meter. Kirchhoff's law directly or mesh analysis, or Nodal analysis, or whatever (Thevenin equivalents, Norton equivalents...). But you need to analyze the circuit appropriately.

The total current supplied by the voltage source is not the current through the meter. You need to find an expression for the current through the meter.
 
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