Calculating power of one of two resistors

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SUMMARY

The discussion focuses on calculating the power dissipated by resistor R2 in a series circuit with resistors R1 and R2 connected to a 3V battery. Given R1 = 10 Ω and R2 = 5 Ω, the total resistance is Rtotal = R1 + R2 = 15 Ω. The correct current through the circuit is I = ∆V / Rtotal = 3V / 15Ω = 0.2 A. The power dissipated by R2 is calculated using P = I^2 x R2, resulting in P = (0.2^2) x 5 = 0.2 W.

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Homework Statement


Two resistors R1 and R2 are connected in series to a battery supplying a voltage of ∆V as shown. How much power is dissipated by the resistor R2 in the form of heat? Use the following data: R1 = 10 Ω, R2 = 5 Ω, ∆V = 3 V
http://tinypic.com/r/153tr91/8

2. Homework Equations
Rtotal = R1 + R2
P=V^2/R
P= I^2 x R

The Attempt at a Solution


Rtot = 10 + 15 = 15 ohms
I = 3/15 = .2 A
P = (.2^2) x 10 = .4 WWhy is this the wrong answer??
 
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Because R2 = 5 Ohms, not 10 Ohms
 

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