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Blu3eyes
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Homework Statement
When a 60 [tex]\Omega[/tex] resistor and a 120 [tex]\Omega[/tex] are connected in parallel across a battery there is a current of 700 mA flowing through the battery. When the 120 Ohms resistor is disconnected from the circuit the current through the battery drops to 500 mA. What is the emf and the internal resistance of the battery?
Homework Equations
emf= I(R load + ri)
V terminal = emf - I*ri
The Attempt at a Solution
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