Internal resitance and electromotive force

AI Thread Summary
The discussion revolves around calculating the voltmeter readings in 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 due to no current flow. Upon closing the switch, the voltmeter reading drops to 1.50 V, with an ammeter reading of 1.20 A, indicating current flow through the circuit. The electromotive force (emf) and internal resistance can be determined using the terminal potential difference equations and the observed values. The analysis emphasizes understanding the impact of internal resistance on voltage readings in different circuit configurations.
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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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