Simple voltage drop / sign convention

[Color_of_Cyan]

Homework Statement

Find V_ad and V_eb

Homework Equations

voltage drop, KVL

The Attempt at a Solution

I actually did:

V_ad

-24V - 4V + 6V

= -22V

but this is wrong.I know going the other way

V_ad: 6V + 8V + 12V

= 26V,

because it is only drop, and plus to minus on drops means they are always positive

and it's the correct answer.But it seems for V_ab the real correct way counting the sources was

V_ad: 24V -4V + 6V

= 26VI thought it was different for voltage sources and sign orientation?So when the drop direction is plus to minus (+ to -), the voltage drop is always positive, for both sources AND given sign convention?
Sorry for asking this now. KCL is much easier.
V_eb would be

8V - 6V + 24V

= 10V

 

[lewando]

So when the drop direction is plus to minus (+ to -), the voltage drop is always positive, for both sources AND given sign convention?

For the passive components this is true and for the voltage source you have shown, this is true.

Sometimes you will encounter a "negative voltage source" (a voltage drop moving from + to - terminals which is negative). These are thrown at you, occasionally, to see how well you react.

V_eb would be

8V - 6V + 24V

= 10V

You have a simple math error there, I'm afraid.

The drops encountered moving from e to b clockwise are:

-8V (because the voltage increased resulting in a "negative drop")
-6V (same reason)
+24V (because the voltage dropped by 24V going from a to b)

Summing the drops: -8V -6V +24V = 10 V

 

[256bits]

Keep your notation correct
V[itex]_{eb}[/itex] = -V[itex]_{be}[/itex]

as an example
V[itex]_{af}[/itex] = 6v
V[itex]_{fa}[/itex] = -6v