Internal Resistance problem when batteries are connected in opposition

AI Thread Summary
In a circuit with two batteries connected in opposition, the resultant electromotive force (emf) can be calculated by subtracting the emf of the first battery from the second. Given battery 1 with an emf of 6V and battery 2 with 24V, the resultant emf is 18V. The internal resistances of the batteries and the resistor connecting them are noted but do not affect the basic calculation of resultant emf in this scenario. The key principle is that when batteries are connected in opposition, their emfs effectively subtract from one another. Understanding this rule clarifies how to approach similar problems in electrical circuits.
andipandi
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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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