Find the voltage across the resistor

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SUMMARY

The discussion centers on calculating the voltage across a resistor in a circuit with a 12.0 V battery, internal resistance of 1.0 Ω, and two 6.9 kΩ resistors in series. The ammeter reading was correctly calculated at 1.09 mA, while the initial voltmeter reading of 7.52 V was incorrect. The correct approach involved calculating the equivalent resistance of the voltmeter and the resistor in parallel, leading to the final voltage calculation using V=Itotal*Rparallel.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Knowledge of series and parallel resistor combinations
  • Familiarity with internal resistance in circuits
  • Basic circuit analysis techniques
NEXT STEPS
  • Study the concept of internal resistance in batteries and its effects on circuit performance
  • Learn how to calculate equivalent resistance in parallel and series circuits
  • Explore the use of ammeters and voltmeters in circuit measurements
  • Investigate advanced circuit analysis techniques, such as Thevenin's and Norton's theorems
USEFUL FOR

Students studying electrical engineering, hobbyists working on circuit design, and anyone interested in understanding voltage measurements in resistor networks.

ooohffff
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Homework Statement


A battery with
scriptE.gif
= 12.0 V and internal resistance r = 1.0 Ω is connected to two 6.9 kΩ resistors in series. An ammeter of internal resistance 0.50 Ω measures the current, and at the same time a voltmeter with internal resistance 10.0 kΩ measures the voltage across one of the 6.9 kΩ resistors in the circuit. What do the ammeter and the voltmeter read?

Homework Equations


V=IR

The Attempt at a Solution


So I found the ammeter reading where Itotal=1.09mA which was correct.
For the voltmeter reading, I got:
V=Itotal*6.9kΩ=7.52V which was incorrect. Anyone care to help?
 
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ooohffff said:
Anyone care to help?
Help what? You haven't shown any work, just answers. That's not the way it works here.
 
phinds said:
Help what? You haven't shown any work, just answers. That's not the way it works here.
Ammeter reading:
Rparallel=[(1/6.9k)+(1/10k)]-1=4082.84
Rtotal= 6.9kΩ+4082.84Ω+1Ω+.5Ω=10984.34Ω
Itotal=V/R=12/10984.34=.00109A

Better?
 
Ah, I got it it's V=ItotalRparallel
 

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