Voltage in a Circuit: Find Vx with 2A and 6Ω

[Schaus]

Homework Statement

Determine the voltage, Vx, if 2A passes through the 6 Ω resistor.

Homework Equations

V = IR

The Attempt at a Solution

1/Req=2/12Ω + 1/12Ω = 3/12Ω
Reciprocal of 3/12Ω = 12/3 = 4 Ω

I=2A
R=6 Ω
V = IR
V = (2A)(6 Ω)
V = 12V

Total resistance is 4 Ω + 3 Ω + 2 Ω = 9 Ω
The answer is supposed to be 27V but I don't know how to get it.

 

[Merlin3189]

So far so good!
You know the voltage across the 6Ω and 12Ω resistors. You therefore know or can calculate the current through each and the total current from the source.

If you know the current from the source, then you know the current through the other two resistors and can calculate the voltage across them.

Then you know all the voltages and can work out the source voltage.

 

[berkeman]

Homework Statement

Determine the voltage, Vx, if 2A passes through the 6 Ω resistor.

Homework Equations

V = IR

The Attempt at a Solution

1/Req=2/12Ω + 1/12Ω = 3/12Ω
Reciprocal of 3/12Ω = 12/3 = 4 Ω

I=2A
R=6 Ω
V = IR
V = (2A)(6 Ω)
V = 12V

Total resistance is 4 Ω + 3 Ω + 2 Ω = 9 Ω
The answer is supposed to be 27V but I don't know how to get it.

Welcome to the PF.

I would approach this problem a little differently. Since we are given 2A through the 6 Ohm resistor, that gives you the 12V number that you got in your work. But that then gives you the current through the 12 Ohm resistor too, right? So what is the total current flowing around the loop?

Then use the parallel combination of the 6 Ohm and 12 Ohm resistors plus the other series resistors and that current to tell you the source voltage. Makes sense?

 

[Merlin3189]

I agree with Berkeman. I was just trying to follow through with your approach.

 

[berkeman]

I agree with Berkeman. I was just trying to follow through with your approach.

Either way should work. I just like to look for the easiest way to approach a circuit problem, where possible. Many times there is no simple way...

 

[Schaus]

So you're saying I take the 12V from the 6 Ohm and 12 Ohm. Then take (2A)(3Ohm) + (2A)(2Ohm)? Because that would be 12V + 6V + 4V = 22V and the answer I am given is 27V

 

[berkeman]

So you're saying I take the 12V from the 6 Ohm and 12 Ohm. Then take (2A)(3Ohm) + (2A)(2Ohm)? Because that would be 12V + 6V + 4V = 22V and the answer I am given is 27V

I'm not following what you are saying. The 2A through the 6 Ohms means there is 12V across both the 6 Ohm and 12 Ohm resistors. What does that make the total current through the parallel combination of those two resistors?

 

[Schaus]

I'm not following what you are saying. The 2A through the 6 Ohms means there is 12V across both the 6 Ohm and 12 Ohm resistors. What does that make the total current through the parallel combination of those two resistors?

Ok if I have this straight then I can use I=V/R I = 12V/18Ohms = 0.6667 A? If so I am still unsure of how I determine the rest of the voltage.

 

[berkeman]

12V/18Ohms

Where did that come from? The 12 Ohm and 6 Ohm resistors are in parallel.

What is the current through a 12 Ohm resistor that has 12V across it?

 

[Schaus]

12V/12Ohm = 1A. This gives me 3A. Does this mean I have 3A throughout the whole circuit? Then I can take V = IR V = (3A)(9Ohm) V = 27V

 

[Merlin3189]

How about putting your results into the diagram?
You found a voltage. You found an equivalent resistance for the 6 and 12
Then you found an equivalent resistance for the whole circuit.
You can transfer voltages and currents using Kirchoff's ideas.

 

[Schaus]

Ya I should've done that! Makes perfect sense now! Thanks a lot for the help!

 

[berkeman]

Does this mean I have 3A throughout the whole circuit?

3A)(9Ohm) V = 27V

Yes.