Voltage-divider and d'Arsonval voltmeter problem

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SUMMARY

The discussion centers on the analysis of a voltage-divider circuit using a d'Arsonval voltmeter with a sensitivity of 200 ohms/V and a full-scale rating of 150 V. The no-load output voltage is determined to be 100.8 V when the voltmeter is placed across a 60 kohms resistor. The resistance of the voltmeter is calculated to be 30 kohms. The participants confirm that the voltmeter reading across the 126 V source is indeed 126 V, while the readings across the resistors do not add up to the source voltage due to the loading effect of the voltmeter.

PREREQUISITES
  • Understanding of voltage-divider circuits
  • Knowledge of d'Arsonval voltmeter specifications
  • Familiarity with Ohm's Law and circuit analysis
  • Ability to perform calculations involving resistors in series and parallel
NEXT STEPS
  • Study the voltage-divider equation in detail
  • Learn how to calculate the loading effect of measuring instruments
  • Explore the characteristics and applications of d'Arsonval voltmeters
  • Investigate equivalent circuit analysis for complex resistor networks
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Electrical engineers, physics students, and anyone involved in circuit design or instrumentation who seeks to understand voltage measurement techniques and the impact of meter resistance on circuit behavior.

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Member advised to use the homework template for posts in the homework sections of PF.
The voltage-divider circuit is designed so that the no-load output voltage is 8/10ths of the input voltage. A d'Arsonval voltmeter having a sensitivity of 200 ohms/V and a full-scale rating of 150 V is used to check the operation of the circuit.
a) What will the voltmeter read if it is placed across the 126 V source?
b) What will the voltmeter read if it is placed across the 60 kohms resistor?
c) What will the voltmeter read if it is placed across the 15 kohms resistor?
d) Will the voltmeter readings obtained in parts (b) and (c) add to the reading recorded in part (a)? Explain why or why not.

Output voltage is 100.8 V. About d'Arsonval voltmeter: using sensitivity and full-scale rating we can find resistance of the voltmeter? Than it is 30 kohms. I think in part (a) answer is 126 V, is it correct? Have some problems with parts (b)-(d). Thanks for sharing your thoughts!
 

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basket_case said:
The voltage-divider circuit is designed so that the no-load output voltage is 8/10ths of the input voltage. A d'Arsonval voltmeter having a sensitivity of 200 ohms/V and a full-scale rating of 150 V is used to check the operation of the circuit.
a) What will the voltmeter read if it is placed across the 126 V source?
b) What will the voltmeter read if it is placed across the 60 kohms resistor?
c) What will the voltmeter read if it is placed across the 15 kohms resistor?
d) Will the voltmeter readings obtained in parts (b) and (c) add to the reading recorded in part (a)? Explain why or why not.

Output voltage is 100.8 V. About d'Arsonval voltmeter: using sensitivity and full-scale rating we can find resistance of the voltmeter? Than it is 30 kohms. I think in part (a) answer is 126 V, is it correct? Have some problems with parts (b)-(d). Thanks for sharing your thoughts!
Your answer for (a) is correct. Please show us your work on the rest of the parts of the problem. What is the equation for a Voltage Divider?
 
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berkeman said:
Your answer for (a) is correct. Please show us your work on the rest of the parts of the problem. What is the equation for a Voltage Divider?
Voltage divider equation: Uout = Uin * R2 / (R1 + R2). And for my problem Uout = 126 V * 60 kohms / (60 kohms + 15 kohms) = 100.8 V.
 
Your comment about calculating the meter resistance is correct.

Try drawing the equivalent circuit for the set up in b)
 
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CWatters said:
Your comment about calculating the meter resistance is correct.

Try drawing the equivalent circuit for the set up in b)
Here it is.
 

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OK now replace the real world volt meter with an ideal meter and a resistor
 
CWatters said:
OK now replace the real world volt meter with an ideal meter and a resistor
Can I use this equations: U2 = I * R;
I = Uin/(R1 + R2)?
I = 126 V / (15 kohms + 60 kohms) = 1.68 mA; U2 = 1.68 mA * 60 kohms = 100.8 V. Is that correct?
 
Thats the voltage that an ideal meter would display not the voltage this meter would display. See my post #7.
 

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