two identical cells of emf E and internal resistance r are connected in series.
a 7 ohm resistor is connected across the combination and draws a current of 0.333A.
the two cells are now connected in parallel; the 7 ohm resistor now draws a current of 0.375A from the combination.
calculate the emf and internal resistance of the cells.
E - Ir = IR
Sum of potential differences in a closed loop = 0
Current entering a junction = current leaving a junction
The Attempt at a Solution
i managed to get one equation from the series circuit:
2(E - 0.333r) = 0.333 x 7
and an equation for the parallel circuit:
E - 0.1875r = 0.375 x 7
but this didn't give the correct answers from the book, which says that E = 1.5 and r = 1