Terminal Voltage: Solving for 12V Motorcycle Battery Charger

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SUMMARY

The terminal voltage of a 12.0V motorcycle battery with an internal resistance of 0.520 Ω, charged by a current of 11.7 A, is calculated using the formula ΔV = ε - IR. The correct calculation yields a terminal voltage of 5.916V. However, the sign of the current must be considered, as the charger inputs current into the battery, leading to a negative current value. This adjustment is crucial for accurate results, as indicated by the feedback from WebAssign and guidance from the physics teacher.

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  • Understanding of Ohm's Law
  • Familiarity with internal resistance concepts
  • Knowledge of battery charging principles
  • Ability to interpret circuit diagrams
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  • Learn about battery charging circuits and their components
  • Explore advanced circuit analysis techniques
  • Research the implications of current direction in circuit calculations
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Students in physics or electrical engineering, educators teaching circuit analysis, and anyone involved in battery technology or charging systems.

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Homework Statement



(a) Find the terminal voltage of a 12.0-V motorcycle battery having a 0.520 Ω internal resistance, if it is being charged by a current of 11.7 A.

(b) What is the output voltage of the battery charger?


Homework Equations



ΔV = ε - IR
or
Vterminal = ε - IR

The Attempt at a Solution



ΔV = 12 - (11.7*0.52) = 5.916 = Vterminal

This is incorrect. WebAssign also returned this information:

"What is the sign of the current if it is flowing into the positive terminal of the battery? How is terminal voltage related to the internal resistance and battery emf?"

So I made it negative and submitted the response: -5.916

Still wrong. What am I missing?
 
Physics news on Phys.org
You have an external voltage (the "charger") supplying a given current into the battery (which includes an internal resistance). Draw the circuit and pencil in the current and potential drops for the given current.
 
My physics teacher was able to help me. I was missing the fact that the current would have to be made negative since the charger was inputting a current. Thanks though!
 

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