DC Circuit. Kirchoff's 2nd law.

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SUMMARY

The discussion focuses on applying Kirchhoff's Second Law (KSL) to determine the potential difference (P.D.) across a 10-ohm resistor in a DC circuit. The user initially miscalculated the current (I) through the resistor, incorrectly using the equation I(10) + 0.25(12) = 2. Upon correction, the user established that I = 0.5 A by using the equation I(10) - 0.25(12) = 2. The conversation emphasizes the importance of accurately applying KSL to analyze circuits with multiple EMF sources.

PREREQUISITES
  • Understanding of Kirchhoff's Second Law (KSL)
  • Basic knowledge of electrical circuits and components
  • Familiarity with Ohm's Law
  • Ability to analyze series and parallel circuits
NEXT STEPS
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  • Learn about the impact of EMF sources in parallel configurations
  • Explore practical circuit simulation tools like LTspice
  • Investigate common mistakes in circuit analysis and how to avoid them
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Students studying electrical engineering, educators teaching circuit analysis, and hobbyists interested in DC circuit design and analysis.

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Homework Statement



Is it possible to find the P.D. across the 10ohm resistor.

Circuit2.jpg



Homework Equations



Kirchoff's second law.

The Attempt at a Solution



Well KSL states that the sum of the pds is equal to the sum of the emfs.

[itex]I(10) + 0.25(12) = 2[/itex]

Gives the wrong value for I. I is 0.5. I took this part of the circuit out. Actually there is a source of emf in parallel with 5V flowing in the opposite direction to the 2V cell. I'm asking if it's possible to work out the pd across the 10ohm resister without that part of it, using Kirchoffs law for pds and emf in a closed loop.

EDIT: Oops.

[itex]I(10) - 0.25(12) = 2[/itex]

[itex]I = 0.5[/itex]

Sorry.
 
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Why is the value of I wrong :confused:
You have worked out the current I. Which very well gives the p.d.
 

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