How Do You Calculate EMF and Internal Resistance in a DC Circuit?

[natural]

Homework Statement

To find the emf and internal resistance of an unmarked battery you are given two resistor of value 5.0(ohms) and an ammeter. When the resistors are connected in parallel the current drawn from the battery is 2.0A. What are the e.m.f and internal resistance of the battery?

Am i going right?

Homework Equations

E = IR = Ir

The Attempt at a Solution

Series
R = 5(ohms) + 5(ohms) = 10(ohms)

Parellel
1/R = 1/5 + 1/5 = 2/5
R = 2.5(ohms)

E = 0.75*10 + o.75*r
E = 7.5 + 0.75r ------- (1)

E = 2*2.5 + 2r
E = 5 + 2r ---------- (2)

7.5 + 0.75r = 5 +2r
2.5 = 1.25r
2(ohms) = r

there for r in equation 1 gives
E = 7.5 + 0.75(2)
= 9V

 

[natural]

***TYPO****-*********Relavant Equation should be E = IR + Ir*******

 

[berkeman]

You didn't include the series combination part in your problem statement. But looking at your work, it would appear that you get 750mA when you connect them in series?

If so, then the rest looks correct.