Internal resitance and electromotive force

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SUMMARY

The discussion focuses on analyzing a circuit with a 12.0 V battery, a 4.50 ohm resistor, and an internal resistance of 0.30 ohms. When the switch is open, the voltmeter reads 2.0 V across the battery terminals and 0 V across the resistor. When the switch is closed, the voltmeter reading drops to 1.50 V, and the ammeter indicates a current of 1.20 A. The electromotive force (emf) of the battery is calculated using the formula for terminal potential difference, leading to the determination of the internal resistance as part of the overall circuit analysis.

PREREQUISITES
  • Understanding of Ohm's Law and circuit analysis
  • Familiarity with electromotive force (emf) concepts
  • Knowledge of internal resistance in batteries
  • Ability to interpret voltmeter and ammeter readings in circuits
NEXT STEPS
  • Study the application of Ohm's Law in series circuits
  • Learn about the impact of internal resistance on battery performance
  • Explore the concept of terminal voltage in electrical circuits
  • Investigate the calculation of emf using different circuit configurations
USEFUL FOR

Electrical engineering students, circuit designers, and anyone interested in understanding battery behavior and circuit analysis.

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Homework Statement



1) a complete circuit consists of a 12.0 V battery with a 4.50 ohm resistor and switch. the internal resistance of the battery is 0.30 ohms when switch is open. what does the voltmeter read when placed:

a. across the terminal of the battery when the switch is open
b. across the resistor when the switch is open
c. across the terminal of the battery when the switch is closed
d. across the resistor when the switch is closed

2) when the switch S is open, the voltmeter V reads 2.0 V. when the switch is closed, the voltmeter reading drops to 1.50 V and the ammeter reads 1.20 A. find the emf (electromotive force) and the internal resistance of the battery. assume that the two meters are ideal so they don't affect the circuit.


Homework Equations


Terminal Potential Difference= -Ir+ electromotive force
Terminal Potential Diff= IR
emf=W/q


The Attempt at a Solution


@1 would it be Terminal Voltage= IR? but i kinda don;t know how to apply it with the voltmeter placed in different places.
 
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Just do it the same way as the first one. You can treat the internal resistance as a separate resistor.
 
In each case, you have the following arrangement (all in series)...

Battery positive voltage...internal resistance...switch...load resistor...battery negative.

Bearing in mind that there is no voltage across a resistor if there is no current flowing through it, what would be the open circuit voltage of the battery in each case?

Now, close the switch and consider there are just two resistors (internal resistance and load resistance) in series across the EMF of the battery. The total current would just depend on the total resistance, so what would be the voltage across each resistor?
 

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