Applying Kirchoff's Rule: Solving for Unknown Quantities in a Circuit

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SUMMARY

The discussion focuses on applying Kirchoff's Rule to solve for unknown quantities in a circuit involving a 35.0 V battery, a resistor R1 of 6.00 ohms, and a resistor R2 of 1.00 ohm. The current through the circuit is determined to be 2 A using the sum of currents at a junction. To find the resistance R and the unknown emf ε, participants emphasize the importance of correctly applying Kirchoff's voltage law, which states that the sum of voltage drops around a loop in a circuit must equal zero. The correct voltage values across the components are crucial for accurate calculations.

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  • Understanding of Kirchoff's Laws (current and voltage)
  • Familiarity with Ohm's Law (V = IR)
  • Basic circuit analysis techniques
  • Ability to interpret circuit diagrams
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  • Practice circuit analysis problems involving multiple resistors
  • Learn how to calculate equivalent resistance in series and parallel circuits
  • Explore the concept of electromotive force (emf) in circuits
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Students studying electrical engineering, physics enthusiasts, and anyone looking to deepen their understanding of circuit analysis and application of Kirchoff's Laws.

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Homework Statement


In the circuit shown in fig. 27-29, ε1 = 35.0, R1 = 6.00 , and R2 = 1.00 .

27-29alt.gif


a. Find the current in resistor R.
b. Find the resistance R
c. Find the unknown emf. ε.
d. If the circuit is broken at point x, what is the magnitude of the current in the 35.0 V battery?

Homework Equations


sigmaI=0
sigmaV=0

The Attempt at a Solution


I got that the current is 2 A, based on the sum of currents at a junction. But I can't solve for the resistance of R. I tried V=IR, with I=2, but I guess I was using the wrong V, since 6 is not the answer to b. Then I figured that I needed b to solve for c, so I came here looking for some help. Thanks in advance.
 
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You can (and pretty much have to) use V = IR. As you already figured out, you need to find the correct value of V. Think how you might use Kirchoff's voltage law (that the sum of voltage drops around a loop in a circuit is zero) to do this. Which loop would you pick?

I'd do it by saying that one point in the circuit is at 'zero' voltage (it doesn't matter where you pick) and work out what the voltage of all the other points relative to this (you can use V = IR across resistors - remembering that the V in this equation is the _difference_ in voltage across the resistor) and of course the voltage difference across each battery is the voltage of the battery.

You should be able to work out a voltage for either side of that resistor like that, _then_ you can use V=IR.
 
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