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

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To calculate the emf and internal resistance of an unmarked battery using two 5-ohm resistors and an ammeter, the current drawn when the resistors are in parallel is 2.0A. The calculations involve determining the equivalent resistance for both series and parallel configurations. The equations derived from the circuit analysis yield the internal resistance as 2 ohms and the emf as 9V. A correction was noted regarding the relevant equation for emf, which should include both the current and internal resistance terms. The solution appears accurate based on the provided calculations.
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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
 
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***TYPO****-*********Relavant Equation should be E = IR + Ir*******
 
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.
 
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