Design an Ohmmeter: E & R1 Solutions

  • Thread starter JasonHathaway
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In summary, the conversation discusses designing an Ohmmeter to measure resistances below 110 kilo ohms using an Ammeter with Rm=50 ohms and Ifsd=0.5 mA. The minimum and maximum values of the voltage source that can be used need to be determined, and a scale with at least 5 values needs to be sketched. The controlling resistance, R2, is 2 kilo ohms. The solution involves finding the unknowns E and R1 by assuming R is short circuited and using Kirchhoff's laws to solve for the current in each junction point. It is also suggested to use the current sensitivity, CS, to calculate the unknown resistor.
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JasonHathaway
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Homework Statement


If you an Ammeter with Rm=50 ohms and Ifsd=0.5 mA, What is the voltage differece you should have if you want to design an Ohmmeter to measure the resistances below 110 kilo ohms? Also, determine the values of the minimum and maximum values of the voltage source that can be used, and sketch the scale with 5 values at least (The controlling resistance (R2) is 2 kilo ohms.
figure1-basic-series-type-ohmmeter1.jpg


Homework Equations


R1=E/Ifsd - R2/2 - Rm
CS=1/Ifsd

The Attempt at a Solution


It's the first time I deal with an unknown voltage source, because in this case we have two unknowns: E and R1 (the series resistor).
If E is known, then we assume R short circuit. then we get R1 --> R1=E/Ifsd - R2/2 (the arrow in the middle) - Rm
Then we calculate the unknown resistor in different cases of the current (e.g. 1/3 Ifsd, 1/2 Ifsd, 3/4 Ifsd, 0.95 Ifsd (nearly maximum)).
But in this case - as I said before - there are two unknowns!

I tried to use the following:
I know that the current sensitivity is 1 over Ifsd ---> CS=1/Ifsd=1/0.5mA=10 kilo ohms/V, and Rscale=Vmax/CS.

Does that make any sense?P.S. If anyone have any textbooks or educating sites with similar problem, I'll appreciate putting them here :)
 
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  • #2
I agree with shorting A to B and that Im should then be equal to Ifsd. And it is sensible to start with R2/2. From there you need to follow Kirchhoff: Sum the currents in each junction point and solve (with E as a parameter).
 

FAQ: Design an Ohmmeter: E & R1 Solutions

1. What is an ohmmeter and how does it work?

An ohmmeter is a device used to measure the resistance of an electrical circuit. It works by sending a small known current through the circuit and then measuring the voltage drop across the circuit. The resistance can then be calculated using Ohm's law (R=V/I).

2. What are the components needed to design an ohmmeter?

The main components needed to design an ohmmeter are a power source, a known resistance (R1), a voltmeter, and a switch. The power source provides the current, R1 is used to create a known voltage drop, the voltmeter measures the voltage drop, and the switch is used to control the flow of current through the circuit.

3. How do you calculate the resistance of a circuit using an ohmmeter?

The resistance can be calculated by using the formula R=V/I, where V is the voltage drop across the circuit and I is the current flowing through the circuit. The voltage drop can be measured using the voltmeter and the current can be calculated by dividing the known voltage drop (created by R1) by the resistance of the circuit.

4. What are the limitations of using an ohmmeter?

One limitation of using an ohmmeter is that it can only measure the resistance of a circuit when it is not connected to a power source. If the circuit is powered on, the ohmmeter may give inaccurate readings or even get damaged. Another limitation is that an ohmmeter can only measure resistance and cannot be used to measure other electrical properties such as voltage or current.

5. How can the accuracy of an ohmmeter be improved?

The accuracy of an ohmmeter can be improved by using a more precise voltmeter and a known resistance (R1) with a smaller tolerance. The circuit should also be designed with low resistance connections to minimize any additional resistance in the circuit. Regular calibration and maintenance of the ohmmeter can also help improve its accuracy.

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