Solving for EMF: Potential Differences with R1 = 4 x R2 | Homework Question

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SUMMARY

The discussion focuses on calculating the potential difference in a circuit where R1 equals four times R2, specifically addressing the relationship between EMF (E) and the resistances. The user initially assumed R1 = 3/4 and R2 = 1/4 but later derived the correct values using Ohm's Law (V = IR) and the equation for EMF (E = v1 + v2). The final expressions for current (I) and EMF across R2 were derived, emphasizing the importance of substituting R1 in terms of R2 to simplify calculations.

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Homework Statement


If R1 = 4 x R2 then what are the values of the potential difference in terms of the EMF, E?
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I have R1 = 3/4, and R2 = 1/4 but that seems a bit simplistic, then I tried plugging it into equations and got completely different answers.

Homework Equations


R=V/I (?)
EMF=v1 +v2


The Attempt at a Solution


First I simply put it as 1/4 and 3/4, then I plugged it into the equations:
v=IR
v1=1(4) =4
v2=1(1) =1
EMF=v1 + v2 = 5
R2 = 1.25, R1 = 5

??
 
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Solve the current, I. Express this interim result as a function of E, R1 and R2.

Since you know that R1 = 4 x R2, you can now express the current solely as function of E and R2 (since R1 goes away after the substitution).

Now that you know the current, using V = IR, find the EMF (i.e. voltage) across R2. You should be able to express it solely as a function of E (after some algebraic simplification).
 

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