How Do You Calculate the Resistance of a Voltmeter?

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SUMMARY

The resistance of a voltmeter connected to a 5.0 V battery, which reads 4.9 V, is calculated to be 24.5 Ω. The internal resistance of the battery is determined to be 0.5 Ω using the formula E = I (r + R), where E is the EMF, I is the current, r is the internal resistance, and R is the load resistance (12 Ω). The current through the voltmeter is 0.2 A, derived from the equation 4.9 = 5 - I r. This analysis clarifies that the current of 0.40 A applies only when a 12 Ω resistor is connected, not when the voltmeter is in use.

PREREQUISITES
  • Understanding of Ohm's Law and its applications
  • Familiarity with the concept of internal resistance in batteries
  • Knowledge of circuit analysis techniques
  • Ability to manipulate algebraic equations for electrical calculations
NEXT STEPS
  • Study the effects of internal resistance on battery performance
  • Learn about voltmeter calibration techniques
  • Explore advanced circuit analysis methods, such as Thevenin's theorem
  • Investigate the impact of load resistance on voltage readings
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This discussion is beneficial for electrical engineering students, circuit designers, and anyone involved in battery testing and measurement techniques.

moenste
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Homework Statement


A battery is known to have an EMF of 5.0 V but when a certain voltmeter is connected to it the reading is 4.9 V. The battery can deliver a current of 0.40 A when connected to a resistance of 12 Ω. What is the resistance of the voltmeter?

Answer: 24.5 Ω.

2. The attempt at a solution
Let's find the internal resistance first: E = I (r + R) → 5 = 0.40 = (r + 12) → r = 0.5 Ω.

Then we have 4.9 = 5 - I r → I = 0.2 A and the resistance is R = V / I = 4.9 / 0.2 = 24.5 Ω.

Is this correct? I am mostly unsure why do we look for current in here 4.9 = 5 - I r. The internal resistance is in series with the battery and the current is equal to 0.4 A.
 
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moenste said:
Is this correct? I am mostly unsure why do we look for current in here 4.9 = 5 - I r. The internal resistance is in series with the battery and the current is equal to 0.4 A.
The current is not 0.4 A when it's just the meter attached to the battery. The 0.40 A occurs when the 12 Ω resistor is connected to the battery. They are two separate experiments.

upload_2016-10-3_10-9-16.png
 
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gneill said:
The current is not 0.4 A when it's just the meter attached to the battery. The 0.40 A occurs when the 12 Ω resistor is connected to the battery. They are two separate experiments.

View attachment 106878
Case 2 is the original case, right?
 
moenste said:
Case 2 is the original case, right?
What do you consider to be the original case?
 
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gneill said:
What do you consider to be the original case?
E = 5 V, I = 0.4 A, R = 12 Ohm.
 
moenste said:
E = 5 V, I = 0.4 A, R = 12 Ohm.
Then yes, that's what's depicted in Case 2.
 
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A simplifying assumption was made that battery source impedance is invariant with respect to current.
 
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