Circuits Q&A: My Question About Vdb

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The discussion centers on the relationship between Vbd and Vdb in the context of charge flow. It asserts that Vbd equals VR2, indicating that current from point d to point b gains potential energy. However, the original poster questions this interpretation, noting that electrons, which flow opposite to conventional current, lose potential energy when moving from lower to higher potentials. This leads to the conclusion that Vbd should actually equal -VR2. Clarification is sought on this apparent contradiction regarding the behavior of electrons and potential energy.
Miike012
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My question is regarding Vdb.
for pos flow of charge

Vbd = -Vdb
and
Vdb = -VR2
so Vbd = VR2.

This means that the current flowing from point d to point b gained potential energy. However the flow of charge from d to b are electrons and electrons lose potential energy as they move from lower to higher potentials. So when I look at it in that point of view
Vbd should be equal to -VR2.

Can someone clarify to me why I am wrong.
 

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Electrons are negatively charged, and flow in the opposite direction of the current.
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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