Engineering Find the Voltage in the following circuit diagram

[vaio911]

Homework Statement

Find V_x in the circuit.

Homework Equations

\displaystyle{\sum V_i=0}

The Attempt at a Solution

I tried to add all of the voltages and label the 2 ohm res V1 and 1 ohm res V2.

Then I did:

V_x+V_1-15+V_2+2V_x=0

3V_x+V_1+V_2=15

but then I couldn't find any other equations that will aid me in solving this.
I'm not really sure if I really need the V1 and V2.
The circuit doesn't give you the current either, so I don't know why resistance would help.

Thanks.

 

[berkeman]

Homework Statement

Find V_x in the circuit.

Homework Equations

\displaystyle{\sum V_i=0}

The Attempt at a Solution

I tried to add all of the voltages and label the 2 ohm res V1 and 1 ohm res V2.

Then I did:

V_x+V_1-15+V_2+2V_x=0

3V_x+V_1+V_2=15

but then I couldn't find any other equations that will aid me in solving this.
I'm not really sure if I really need the V1 and V2.
The circuit doesn't give you the current either, so I don't know why resistance would help.

Thanks.

Welcome to the PF.

Where did you come up with those equations? There is one current around that loop that generates the voltage drops and controls the voltage controlled current source. Could you please try writing the one equation for the loop again, and solving it?

 

[vaio911]

reply

Is this right?

i is the current

\begin{align*}<br /> 2V_x+V_x+2i+i&amp;=15\\<br /> 3(5\cdot i)+2i+i&amp;=15\\<br /> i&amp;=\frac{15}{18}=\frac{5}{6}\;A<br /> \end{align*}

V=I\cdot R=\frac56 \cdot 5=\boxed{\dfrac{25}{6}\;V}

 

[vaio911]

Also, current is the same throughout in a series circuit right/

 

[berkeman]

Is this right?

i is the current

\begin{align*}<br /> 2V_x+V_x+2i+i&amp;=15\\<br /> 3(5\cdot i)+2i+i&amp;=15\\<br /> i&amp;=\frac{15}{18}=\frac{5}{6}\;A<br /> \end{align*}

V=I\cdot R=\frac56 \cdot 5=\boxed{\dfrac{25}{6}\;V}

The diamond source is a voltage controlled current source, right? Not a voltage...

Also, current is the same throughout in a series circuit right/

Correct.

 

[berkeman]

Upon further review, and input from another Mentor, I may be wrong about the controlled source. It looks like a current source to me, but I could be wrong about that. It may be a voltage controlled voltage source (VCVS) instead. Sorry if I mislead you on this.