Charge moves through a circuit element and loses energy

AI Thread Summary
To find the voltage across the circuit element when a +6 C charge loses 12 J of energy, the correct calculation is V = J/C, resulting in 2 V. The charge experiences a voltage drop as it loses energy while passing through the circuit element. The initial calculation mistakenly used 2 C instead of 6 C, which should be corrected to V = 12 J / 6 C = 2 V. The reasoning about the voltage drop is accurate, as energy loss indicates power absorption by the element. Overall, the approach to the problem is sound, but attention to detail in calculations is crucial.
Westin
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Homework Statement



If +6 (C) charge moves through a circuit element, and in so doing the charge loses 12 (J) of energy, then what is the magnitude of the voltage across this circuit element?

A)2(V)
B)6(V)
C)12(V)
D)72(V)

For the same circuit element and situation described in question 1, is the +6 (C) of charge passing through a voltage rise, or a voltage drop?

A) voltage rise
B) voltage drop
C) none of the above

2. Homework Equations

V = J/C[/B]

The Attempt at a Solution



V = 12(J)/2(C) = 2(V)

Since the circuit lost energy, it will be a voltage drop because the element absorbed power.
[/B]
Is my reasoning right here? It seems too easy for it to be so simple.

Thanks
 
Last edited:
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Westin said:
V = 12(J)/2(C) = 2(V)
Typo [6(C)?]. Otherwise, looks OK. You can check with a basis of 1 sec then using J, V, A, and W equations.
 
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