A battery-battery-resistor circuit

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SUMMARY

The discussion centers on calculating the internal resistance of batteries in a circuit consisting of two 1.5V batteries connected in series with an 18 Ohm lamp. The circuit produces 0.25 Watts to the lamp, leading to the conclusion that the internal resistance of the batteries must be considered. The correct approach involves using the power formula P=VI and Ohm's law to derive the internal resistance, resulting in a total internal resistance of 18 Ohms after accounting for the lamp's resistance.

PREREQUISITES
  • Understanding of Ohm's Law (V=IR)
  • Familiarity with power calculations (P=VI)
  • Basic knowledge of series circuits
  • Concept of internal resistance in batteries
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  • Study the impact of internal resistance on circuit performance
  • Learn about power distribution in series circuits
  • Explore advanced circuit analysis techniques
  • Investigate different battery types and their internal resistances
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Students studying electrical engineering, hobbyists building circuits, and anyone interested in understanding battery performance in practical applications.

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Homework Statement


A circuit:
Battery (1.5v) -- Battery (1.5v) -- Lamp(18 Ohm) and back to the first battery.

The circuit is producing 0.25 Watt to the lamp. what is the inside-resistance of the batteries?

Homework Equations


P=VI
IR=V

The Attempt at a Solution


1.5+1.5=3Volt
P=0.25Watt
0.25=3*I
I=1/12
(1/12)*R=3
R=36
we already have 18 Ohms from the lamp so the remining 36-18=18 is the answer.
WRONG:confused:
 
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You cannot assume that the voltage across the lamp is 3V--not when the batteries have internal resistance. Instead, find an expression for power consumption in the lamp in terms of I and R, and solve for the current.
 
Last edited:
Oh, I got it now, Thanks!
 

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