Solving for current in a circuit

[Miike012]

I have no idea what I a doing. I don't know how to deal with the two symbols (the one that looks like a diamond and a circle with an arrow).

If those were replaced with a battery I would be able to solve it, but I can't solve it because I don't know which direction the current is flowing in each branch of the circuit.

Anyways I did KVL for each loop and can someone tell me if I am correct?

Loop 1:
I₀(6kΩ) - V_x + i1(8Ω) = 0
Loop 2:
V_{voltage over element of 5A} - i1(8Ω) + V_x = 0
Loop3: -3V_x - V_x = 0

Out most loop: i₀ - 3V_x = 0

Can someone let me know if that is correct?

Then what should I do next to solve for i₀?

 

[Miike012]

I just need to know which direction the current is going through each branch. does anyone know?

 

[gneill]

I have no idea what I a doing. I don't know how to deal with the two symbols (the one that looks like a diamond and a circle with an arrow).

If those were replaced with a battery I would be able to solve it, but I can't solve it because I don't know which direction the current is flowing in each branch of the circuit.

Anyways I did KVL for each loop and can someone tell me if I am correct?

Loop 1:
I₀(6kΩ) - V_x + i1(8Ω) = 0
Loop 2:
V_{voltage over element of 5A} - i1(8Ω) + V_x = 0
Loop3: -3V_x - V_x = 0

Out most loop: i₀ - 3V_x = 0

Can someone let me know if that is correct?

Then what should I do next to solve for i₀?

The circle with the arrow inside is an ideal current source. It will always inject exactly 5 A no matter what. The diamond with the arrow inside is called a controlled current source. It will produce 3V_x amps. That is, by some means not shown it "measures" the potential V_x across the 4 Ω resistor and then produces an amperage of 3 times the magnitude of that potential difference. If you wish, you can think of the coefficient "3" as having units of Ohms so then it becomes 3 Ω * V_x, which by Ohm's law results in Amperes.

Regarding your solution attempt, is there some particular reason you chose to use KVL loop analysis rather than nodal analysis? The reason I ask is that your circuit contains only current sources and one independent node, which makes it very amenable to nodal analysis.

If you really, really want to use loop analysis, since the two current sources are in parallel I'd suggest combining them into a single controlled current source: I = 5 - 3V_x amps directed upwards. That will eliminate the third loop entirely.

When you first write your loop equations, don't use V_x as a potential drop. Write it as -I1*4Ω. Use that for V_x everywhere; that will tie in the current I1 to the controlled source's current in your equations without needing a separate equation for V_x.

 

[gneill]

I just need to know which direction the current is going through each branch. does anyone know?

You won't know for sure until you've solved the circuit. If you guess wrong for a current at the outset, the value you find will be negative. No big deal, that just means it's really flowing in the opposite direction to your guess.

 

[Miike012]

edit..