The emf of a battery A is balanced by a length of 75 cm on a potentiometer wire. The emf of a standard cell, 1.02V, is balanced by a length of 50 cm. What is the emf of A?
Calculate the new balance length if A has an internal resistance of 2 Ω and a resistor of 8 Ω is joined to its terminals.
The Attempt at a Solution
I can answer the first part of this question. Emf of A is 1.53 V. It's the second part of the question that I don't know.
With the addition of the 8 Ω resistor, the voltage across the potentiometer wire must be equal to the voltage from A after voltage drop across 8Ω resistor, right?
And if so, I don't know how to find it, for I don't know how to find the value of current. If my reasoning is correct, I will end up with E= I(R+r) + V', where V' is the potential difference across potentiometer wire.
I don't know if I'm totally wrong.