Circuit question with two batteries and resistors in pa

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SUMMARY

The discussion focuses on calculating the power dissipated by resistor R1 in a circuit with two batteries and resistors. The user applied Ohm's Law (V = IR) and the power formulas (P = VI, I²R, V²/R) to analyze the circuit. The user simplified the bottom resistor pair to 10/7 Ohms and calculated the current I3 as 1.05A. However, confusion arose regarding the current through R1, leading to the conclusion that the potential difference (P.D.) across R1 is crucial for determining the current flowing through it.

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Homework Statement


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Q: What is the power dissipated by R1?

Homework Equations



Ohm's law: V = IR
Power: P = VI, I2R, V2/R
Kirchoff's circuit and voltage laws

The Attempt at a Solution



I assumed the currents were I1 traveling rightwards from the top battery, I2 rightwards from the second battery, and came together to make I3 for the bottom branch. The bottom resistor pair I simplified to 10/7 Ohms. From here I assume I have to set up simultaneous equations to solve for I1.

I used the equation for the bottom branch 1.5V - 10/7 * I3 = 0 to get I3 = 1.05A.

I subbed this into the equation for the outer loop 1.5 - 3I1 - 10/7I3 = 0. However plugging 1.05A into this gave me 0 for I1...

I feel like I'm way off the mark..
 
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Why would there be any current flowing in R1?
 
oz93666 said:
Why would there be any current flowing in R1?
Indeed!

The current through R1 is determined by the PD across R1. So ask yourself: what is the P.D. across R1?
 
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