Node voltage w/ voltage source

  • Thread starter skyfire101
  • Start date
  • #1
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
 

Answers and Replies

  • #2
vk6kro
Science Advisor
4,081
40
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.
 
  • #3
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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  • #4
vk6kro
Science Advisor
4,081
40
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
 

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