DC Circuit. Kirchoff's 2nd law.

AI Thread Summary
The discussion revolves around applying Kirchoff's Second Law to determine the potential difference (P.D.) across a 10-ohm resistor in a circuit. The user initially struggles with the calculations involving the current (I) and the effects of an opposing emf source. After some corrections, they establish that the current I is 0.5 A, which leads to the correct P.D. across the resistor. The user seeks clarification on why their initial value for I was incorrect. Ultimately, they confirm that the calculations yield the necessary P.D. using Kirchoff's law.
Gregg
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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.

I(10) + 0.25(12) = 2

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.

I(10) - 0.25(12) = 2

I = 0.5

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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