# Node voltage w/ voltage source

when does the voltage of a node with a branch attached to a voltage source not equal the voltage of the voltage source?
This question has been driving me crazy because i am unsure when i am able to apply voltage division to portions of a circuit i try to analyze.
I know the potiential difference across a voltage source always needs to be equal to the value of the voltage source but, if the branch coming out of a voltage source is not at the same potiental as the voltage source how can i apply voltage division?
Any help is much appreciated, thank you

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vk6kro
Can you draw a diagram?

You can usually apply voltage division if you have resistors in series across a voltage source, provided there is no other power source in the circuit.

However, more complex arrangements mean you have to use analysis to get the voltages.

I'm really just looking for a way to apply to apply the rule in general, or to portions of a circuit. I know that voltage division is used mostly for resistances in series, however there are ways to compensate for parallel resistances to. So i was wondering to what scope voltage division applies. My txtbook has examples where it applies it to portions of a circuit, but doesn't explain why it works in said examples.

Is it accurate so say that voltage division still applies to a closed loop even when one branch leads to some interface or output?
In my attachment is the circuit of suspect, it asks to find Vo in terms of input Vs, I can only match the equation they get by using voltage division on the loop containg Vs
i.e
Ix*Rp = Vs*Rp/(Rs+Rp)
(Rp on left gets canceled)

and then sub Ix into another equation that uses volt division on the loop with a dependent source.
i.e
Vo = (-rIx)*RL/(RC + RL)
(sub previous eqn into Ix)
*above Vo should be "+" and below Vo "-" sign

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vk6kro
You can get the current in the left circuit with Vs / (Rs + Rp). call this current I1

The current source gives this current times r

So this current flows in Rc and RL.

The voltage drop across RL is then I1 * r * RL

Substituting for I1
The output voltage = {Vs / (Rs + Rp)} * r * RL