A battery of EMF E and internal resistance r is connected across a variable resistor. When the resistor is set at 2lΩ the current through it is 0.48A; when it is set at 36Ω. the current is 0.30A. Find E and r.
"I1=0.48A; I2=0.3A; R1=21\\Omega; R2=36\\Omega."
"E" -EMF and "r" - internal resistance .
Let's use Ohm's law for whole circuit:
"I=E\/(R+r);" In our case: