a battery with unknown internal resistance and open circuit potential (emf) is connected toa 3 ohm external resistor. the terminal potential is found to be 60 volts. when the same battery is connected to a 48 ohm resistor, the terminal potential is 96 volts. what is the internal resistance of the battery?
electric potential V = IR where I is current, R is resistance
current I = epsilon/R where epsilon is emf
The Attempt at a Solution
I_1 = V_1/R_1 = 60/3 = 20 ampere
I_2 = V_2/R_2 = 96/48 = 2 ampere
now that is as far as i got, to find R internal, do i use a similar formula to R = V/I where V and I are the change in them?
how does the emf come into play?