in a circuit two resistors each of 2 Ω is connected in parallel to a cell such that a current of 1.2 Amps flows through it now when those same two resistors are connected in series a current of 0.4 Amps is detected in the ammeter.(taking resistance of wire as zero)calculate
1)internal resistance of cell
2)EMF of cell
GOOD LUCK!!!!!!
I suspect there are different understandings of the English -- subject verb agreement and such. This is how I understood the description:
R3=battery internal resistance. The current is current through the battery. How about posting your answer hidden as spoiler? Since ammeter resistance wasn't mentioned, I assumed one was not in the circuit. That is, it was a Hall or similar type ammeter.
Les Jones My solution was very similar to yours. I just wrote the equations for each case, equated them, and solved for one variable. Then substituted to get the other variable.
I would have shown my work, but I am terrible with LaTex.