Battery’s terminal potential difference

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SUMMARY

The terminal potential difference of a 1.5 V AA battery with an internal resistance of 1 ohm connected to a total series resistance of 50 ohms is calculated to be 1.47 V. The current flowing through the circuit is determined to be 29.412 mA using the formula I = EMF / (R + r). The solution is confirmed as correct, and an alternative method using the potential divider rule is suggested for further understanding.

PREREQUISITES
  • Understanding of Ohm's Law
  • Familiarity with series circuits
  • Knowledge of internal resistance in batteries
  • Basic concepts of voltage dividers
NEXT STEPS
  • Study the application of Ohm's Law in different circuit configurations
  • Learn about the potential divider rule in detail
  • Explore the effects of internal resistance on battery performance
  • Investigate series and parallel circuit analysis techniques
USEFUL FOR

Students studying electrical engineering, hobbyists working with batteries, and anyone interested in understanding battery performance in circuits.

zohaibtarar
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Homework Statement


An external circuit with a total series resistance of 50 (omega) is connected to a 1.5 V AA battery which has an internal resistance of 1 (omega). What is the battery’s terminal potential difference?

Homework Equations


I = EMF / (R + r)

The Attempt at a Solution


I = EMF / (R + r) = 1.5V/ 51 Ohm = 29.412 mA

V = 1.5V - 0.029412A * 1 Ohm = 1.47 Voltbut i m still confused if i did it right or wrong please help

Thank you!
 
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