Achieving the Balance Point: Current Flow & Galvanometer Reading

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SUMMARY

The discussion centers on achieving the balance point in a circuit involving a galvanometer and resistors. When the balance point is reached, there is no current passing through the galvanometer, indicating a potential difference of zero across it. The calculations provided confirm that the voltage across the 20-ohm resistor is indeed zero, while the voltage across the circuit elements is accurately calculated as 2.7V. The correct application of the voltage divider rule is demonstrated with the equations V_xy = (8.1/120) * 40 and E" = 14.58V.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with voltage divider circuits
  • Basic knowledge of galvanometer operation
  • Ability to perform circuit analysis using Kirchhoff's laws
NEXT STEPS
  • Study the principles of galvanometer calibration and usage
  • Learn about advanced circuit analysis techniques, including Thevenin's and Norton's theorems
  • Explore practical applications of voltage dividers in electronic circuits
  • Investigate the effects of load resistance on circuit performance
USEFUL FOR

Students studying electrical engineering, educators teaching circuit theory, and hobbyists interested in electronics and circuit design will benefit from this discussion.

werson9339
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Homework Statement


when the balance point is achieved , there're no current passing thru , so Galvanometer should show 0 . Am i right... so there's no current passing thru the 20 ohm resistor . so my working is ( 8.1/120 )(40) = 2.7 V ...
correct me if I'm wrong...

Homework Equations

The Attempt at a Solution


V_xy = V_ac = (9/10) x 10 = 8.1v
k = 8.1/120
V_xy = k (40) = 2.7v

(5/ (5+20+2) ) X E " = 2.7 V

E" = 14.58v[/B]
 

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What is the problem statement ?
so there's no current passing thru the 20 ohm resistor
No. It only means there is no current passing through the galvanometer.
 
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No current passing through the galvanometer means there is no potential difference over the thing. If we call the point between the 5 ##\Omega## and the 20 ##\Omega## resistors point z: VAC = VXZ.

For VAC you have the right result (but perhaps a typo?) : VAC / VAB = 40/120 and VAB = 9V * 9/10.

And for E' (single quote in the drawing) you also have the right result.
 

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