EMF's in series how does this work ?

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In the discussion about the series connection of two cells with different e.m.f.s and internal resistances, a key point is the confusion regarding current flow, as the current through each cell can differ due to their internal resistances. The voltmeter, which has a resistance of 50 ohms, plays a crucial role in measuring the terminal voltage but complicates the circuit analysis. Participants emphasize the importance of applying Kirchhoff's voltage law and considering the internal resistances of the cells. The textbook provides specific values for current and voltage, which some find inconsistent with their understanding of series circuits. Overall, the analysis suggests visualizing the setup as parallel components rather than strictly series to clarify current distribution.
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Homework Statement



A secondary cell having an e.m.f. of 2V and an internal resistance of 1ohm is connected in series with a primary cell having an e.m.f. of 1.5V and an internal resistance of 100ohm the negative terminals of each cell is connected to the positive terminal of the other cell. A voltmeter having a resistance of 50ohm is connected to measure the terminal voltage of the cells. Calculate the voltmeter reading and the current in each cell.

Ans given by book:72.8mA, 34.3mA, 1.93V

Homework Equations


Kirchoff voltage law.


The Attempt at a Solution



I can not understand how 2 cells in series can have a different current ? What concept am I missing ?
 
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The current will change according to which cell voltage you are measuring.

The voltmeter has a resistance of 50R... (that's some "voltmeter"...)
 
Is there a picture for the question? I'm not sure how the voltmeter is connected to measure the terminal voltage.
 
Defennder said:
Is there a picture for the question? I'm not sure how the voltmeter is connected to measure the terminal voltage.
Sorry no picture because the textbook has no picture only the words
 
zeitghost said:
The current will change according to which cell voltage you are measuring.

The voltmeter has a resistance of 50R... (that's some "voltmeter"...)
The book answer gives one voltage and 2 currents.
 
Instead of thinking about 2 batteries in series, try picturing the situation as 3 objects all in parallel (the 2 cells and the 50-ohm resistor).

Don't forget to include the internal resistances as part of the cells.
 

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