Internal Resistance problem when batteries are connected in opposition

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SUMMARY

The discussion centers on calculating the resultant electromotive force (emf) in a circuit with two batteries connected in opposition. Battery 1 has an emf of 6V and an internal resistance of 1Ω, while Battery 2 has an emf of 24V and an internal resistance of 0.5Ω. The resistor connecting the batteries is 4.5Ω. The resultant emf is determined to be 18V by subtracting the emf of Battery 1 from that of Battery 2, illustrating the principle that in circuits with opposing batteries, the total emf is the difference between the individual emfs.

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A simple circuit with 2 batteries and a resistor connected in series. The positive terminals are connected to each other via a 4.5 resistor. Their emfs and internal resistances are below. What is the resultant emf?
battery 1: ε=6V r=1Ω
battery 2: ε=24V r=0.5Ω
resistor connecting the two batteries is 4.5Ω


(I've attached a picture of the circuit diagram)

The answer is 18V. I know that resultant emf =ε - Ir. But I have no idea how to achieve an answer when their are two batteries present, let alone connected in opposition. I've spent the past two hours research books and the internet with nothing to show of it. The only way I can see to get 18V would be to simply minus emf of battery 1 from battery2, if this is correct rather than a coincidence, what is the rule behind it?
 

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welcome to pf!

hi andipandi! welcome to pf! :smile:
andipandi said:
The only way I can see to get 18V would be to simply minus emf of battery 1 from battery2, if this is correct rather than a coincidence, what is the rule behind it?

i've no idea why they tell you all about the resistances :confused:

the emf in a circuit (a loop) is simply the sum of the individual emfs :wink:

(and you can see from the long and short lines that the batteries are "facing" opposite ways, so one of them will be minus the other)
 

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