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Mutaja
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Hi.
I have a problem understanding a question for my electronic engineering course. We have a subject that's directly translated to 'the knowledge of electricity and measurement techniques'.
I do hope this is the correct forum to ask for help at.
Given the following circuit:
See attachment.
Construct a voltmeter with max impact/effect at 10V, when you have a moving coil with the following data: Ifull scale = 1mA, Rm = 2k ohm. What will the voltmeter show when it's applied to the 10k ohm resistance (as shown on the figure), and what will the error be in percentage of ideal voltage because of the load effect?
The sensitivity of the instrument is S = 1/I -> S = Rm/Ufullscale
I don't have a specific attempt at a solution as I'm confused, but this is what I've done.
The current in the circuit before the voltmeter is applied is I = U/R = 100V/105kohm = 0.952mA.
With the voltmeter applied, it will create a 2k ohm resistance in parallel with the 10k resitance making the current in the circuit: I = 100V/55kohm+40kohm+10kohm||2kohm = 1.035mA.
The current that runs through the 10k resistance will be:
10k resistance is 5 times larger than the 2k resitance ->5 times more current will flow through it.
1.035mA/6 = 0.1725mA * 5 = 0.6375mA
The voltmeter surely will show: U = R * I -> 10k ohm * 0.6375mA = 6.375V?
The ideal voltage is 0.952mA * 10k ohm = 9.52V
Error will be 67%.
Now, here's the tricky part. The answer to this exercise is R = 8kohm and error is 47.5%.
I'm totally confused! How can a voltmeter show ohm? Please help me with any input you might have, I've been stuck on this problem for three days now.
And do let me know if I need to clearify anything. Thanks in advance!
I have a problem understanding a question for my electronic engineering course. We have a subject that's directly translated to 'the knowledge of electricity and measurement techniques'.
I do hope this is the correct forum to ask for help at.
Homework Statement
Given the following circuit:
See attachment.
Construct a voltmeter with max impact/effect at 10V, when you have a moving coil with the following data: Ifull scale = 1mA, Rm = 2k ohm. What will the voltmeter show when it's applied to the 10k ohm resistance (as shown on the figure), and what will the error be in percentage of ideal voltage because of the load effect?
Homework Equations
The sensitivity of the instrument is S = 1/I -> S = Rm/Ufullscale
The Attempt at a Solution
I don't have a specific attempt at a solution as I'm confused, but this is what I've done.
The current in the circuit before the voltmeter is applied is I = U/R = 100V/105kohm = 0.952mA.
With the voltmeter applied, it will create a 2k ohm resistance in parallel with the 10k resitance making the current in the circuit: I = 100V/55kohm+40kohm+10kohm||2kohm = 1.035mA.
The current that runs through the 10k resistance will be:
10k resistance is 5 times larger than the 2k resitance ->5 times more current will flow through it.
1.035mA/6 = 0.1725mA * 5 = 0.6375mA
The voltmeter surely will show: U = R * I -> 10k ohm * 0.6375mA = 6.375V?
The ideal voltage is 0.952mA * 10k ohm = 9.52V
Error will be 67%.
Now, here's the tricky part. The answer to this exercise is R = 8kohm and error is 47.5%.
I'm totally confused! How can a voltmeter show ohm? Please help me with any input you might have, I've been stuck on this problem for three days now.
And do let me know if I need to clearify anything. Thanks in advance!
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