Finding Internal Resistance through Circuit Analysis

In summary: 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.
  • #1
Safder Aree
42
1

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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  • #2
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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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?
 
  • #8
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?
 
  • #9
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.
 

Related to Finding Internal Resistance through Circuit Analysis

What is internal resistance in a circuit?

Internal resistance is the resistance within a circuit component, such as a battery or a resistor. It is caused by the material properties and design of the component, and it affects the flow of current in the circuit.

Why is it important to find the internal resistance in a circuit?

Finding the internal resistance allows us to understand how efficient a circuit is and how much power is being lost within the components. It also helps us to accurately predict the behavior of the circuit and make necessary adjustments for optimal performance.

How can internal resistance be determined through circuit analysis?

Internal resistance can be determined by measuring the current and voltage at different points in the circuit and using Ohm's law (V=IR) to calculate the internal resistance. It can also be determined by analyzing the voltage-current relationship of the circuit using a graph.

What factors can affect the internal resistance of a circuit?

The internal resistance of a circuit can be affected by the material properties of the components, the temperature, and the design of the circuit. It can also vary depending on the state of the components, such as the charge level of a battery.

How can the internal resistance of a circuit be minimized?

The internal resistance of a circuit can be minimized by using high-quality components with low internal resistance, keeping the circuit at a stable temperature, and designing the circuit for optimal performance. In some cases, using multiple components in parallel can also reduce the overall internal resistance of the circuit.

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