An external circuit with a total series resistance of 50 ohm is connected to a 1.5 V AA battery which has an internal resistance of 1 ohm. What is the battery’s terminal potential difference?

Question

hartnn · Answer

do you know which equation to use?

zohaibtarar · Answer

not sure but i think its I = EMF / (R + r)

hartnn · Answer

yes thats correct

hartnn · Answer

emf = 1.5
R = 50
r = 1
just plug these in

zohaibtarar · Answer

29.412 mA  thats it? or is there any other step left?

hartnn · Answer

Thats just the current flowing through the circuit,
The voltage drop (PD) across the  battery, can now be found using ohm's law.

V_battery = V-  I * r_battery
r_battery =1
V= 1.5
I= 0.029

zohaibtarar · Answer

so final answer is  1.49 volt? correct?

hartnn · Answer

take 0.029 approximately as 0.03

1.5 - 0.03*1 = 1.5 - 0.03  = 1.47V
thats it! :)

zohaibtarar · Answer

and is there any need of diagram in it? if so can you roughly make one to give me an idea?

Thank you so much

hartnn · Answer

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