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## Homework Statement

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?

## Homework Equations

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?