Voltage division question

Thread Starter

Robert Murch

Joined Nov 2, 2015
43
So I understand how to use voltage division if all my resistors are in series. However, I seem to be struggling understanding how to use voltage division if not all my resistors are in series. I have attached a problem, here is how I tried to solve the problem:

Vx = 4(10)/(1+2+(((1/4)+(1/4))^-1)+(((1/10)+(1/10))^-1)) = 4v

That is I want to say 4*10 divided by (4||4)+(10||10)+2+1

However, according to the answer key the answer is 2v. There must be something I am missing about how this voltage division works. I was under the impression that the general formula was: Vx = Vs(Rn/Rt). Please help.
 

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Jony130

Joined Feb 17, 2009
5,022
But this 4 ohm's resistors are connected in parallel so to find Vx voltage you must first find the equivalent resistance of this two resistors.
Vx = 10V * 2Ω /((4Ω||4Ω)+(10Ω||10Ω)+2Ω+1Ω) = 10V * 2Ω/(2Ω + 5Ω + 2Ω + 1Ω) = 10V * 2Ω/10Ω = 10V * 0.2 = 2V
 

WBahn

Joined Mar 31, 2012
24,854
You basic problem seems to be that you want to throw memorized formulas around without understanding what they mean and what their limitations are.

It's not good enough "to be under the impression" that the general equation for a voltage divider is something. Do you understand where that equation comes from and, more importantly, what the restricts are on the circuit in order for it to be valid?

What would you do if just asked to find the current in the total circuit?

What would you do if just asked to find the current in the top 4 Ω resistor?

Knowing the current in that resistor, what would you do if asked to find the voltage across it?
 

Thread Starter

Robert Murch

Joined Nov 2, 2015
43
But this 4 ohm's resistors are connected in parallel so to find Vx voltage you must first find the equivalent resistance of this two resistors.
Vx = 10V * 2Ω /((4Ω||4Ω)+(10Ω||10Ω)+2Ω+1Ω) = 10V * 2Ω/(2Ω + 5Ω + 2Ω + 1Ω) = 10V * 2Ω/10Ω = 10V * 0.2 = 2V
OH right since the voltage drop for elements in parallel is equal then you have to account for both of them I see now thanks!
 
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