Two cells in series with internal resistors -- calculate the current

In summary, the conversation is about a problem with a circuit involving two batteries in series. The person asking for help is struggling to use the V=IR equation and the answer states that there are two different currents. The responder asks the person to provide a circuit that matches the problem description and explains the importance of displaying the circuit for better understanding.
  • #1
lalallaland
5
0
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 volatage of the cells. Calculate the voltmeter reading and the current in each cell.

Ans given by book:72.8mA, 34.3mA, 1.93V
Relevant Equations
Kirchhoff law
Can somebody explain this please? I don't understand this.
 
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  • #2
Welcome to the PF.
lalallaland said:
Can somebody explain this please? I don't understand this.
How about you draw the circuit as described and attach it to a reply? Label the voltages and resistances in the loop, and show us how you can use the equation V=IR to work toward the answer. (We aren't allowed to provide tutorial help here until you show your best efforts). Thanks
 
  • #3
Hey Berkeman,

Thanks a lot for your answer. I'm really stuck at this problem and it is not leaving my mind. I don't really know how to use V=IR here cause the answer tells me there are two different currents.
 

Attachments

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  • #4
lalallaland said:
the answer tells me there are two different currents.
You are given the resistance of the voltmeter. What does that imply?
 
  • #5
  • #6
However the circuit you supplied does not match that description.

Could you explain this statement? Because I think my circuit matches the description.
 
  • #7
If I am reading your circuit correctly, the negative terminal of the 1.5V cell is correctly connected to the positive of the 2V cell. The negative of the 2V cell is connected to ... Where?
 
  • #8
It's very helpful to actually display text and/or graphics being referred to. Here is the circuit given by OP.
1569731019165.png
 
  • Like
Likes Tom.G

What is the formula for calculating the total current in a circuit with two cells in series?

The formula for calculating the total current in a circuit with two cells in series is:
I = (V1 + V2) / (R1 + R2 + r1 + r2)
Where V1 and V2 are the voltages of the two cells, R1 and R2 are the resistances of the external resistors, and r1 and r2 are the internal resistances of the cells.

How do you calculate the total resistance in a circuit with two cells in series?

The total resistance in a circuit with two cells in series is calculated by adding the external resistances and the internal resistances of the cells.
Rtotal = R1 + R2 + r1 + r2

What is the purpose of using two cells in series in a circuit?

Using two cells in series allows for an increase in voltage, as the voltages of the two cells are added together. This can be useful in applications where a higher voltage is needed, such as in powering electronic devices or in charging batteries.

What happens to the current in a circuit with two cells in series if one of the cells has a higher internal resistance?

If one of the cells has a higher internal resistance, it will cause a decrease in the total current in the circuit. This is because the higher internal resistance will act as a greater resistance in the circuit, reducing the overall flow of current.

Can the internal resistance of a cell be ignored when calculating the total current in a circuit with two cells in series?

No, the internal resistance of a cell cannot be ignored when calculating the total current in a circuit with two cells in series. This is because the internal resistance affects the overall resistance in the circuit, which in turn affects the total current. Ignoring the internal resistance would result in an inaccurate calculation of the current in the circuit.

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