Engineering Thevenin Theorem - Open Voltage and Short Circuit Current

[ohdrayray]

Homework Statement

Homework Equations

Thevenin Theorem, Kirchoff's Voltage Law, mesh analysis

The Attempt at a Solution

I'm kind of confused as to how to approach this question as a whole because I'm getting put off from the two voltage sources.

I was thinking that maybe I could use mesh current analysis so then I'd get:

-10 + 1200(i_{1}) + 3300(i_{1} - i_{2}) + 12 = 0
where i_{2} = 0 since it's within an open circuit, so then
1200(i_{1}) + 3300(i_{1}) = -2
4500(i_{1}) = -2
i_{1} = \frac{-2}{4500} A

Then,
V_{oc} = \frac{-2}{4500} * 3300
V_{oc} = 1.5 V

but that doesn't sound right, so I'm not sure how else to approach it. Thanks for help in advance!

 

[gneill]

There's a single closed loop, as you discovered when you realized that i₂ = 0.

The equation that you've written for loop 1 does not appear to be self consistent in terms of voltage polarities. Voltage rises and drops should always correspond to what occurs while traversing a given component in the direction of the assumed current.

So, suppose that you have the assumed current direction for i₁ as you've shown in the diagram and begin "walking around" the loop at the - terminal of the 10V supply. Can you write the KVL equation?

 

[ohdrayray]

Is it meant to be:
-10 + 1200(i₁) + 3300(i₁) - 12 = 0 ?

 

[gneill]

Is it meant to be:
-10 + 1200(i₁) + 3300(i₁) - 12 = 0 ?

That will do I would have given the terms the opposite signs when writing out the equation, but as long as you're rigidly consistent in your methodology everything will work out.

When I write an equation for a loop from a diagram, after choosing the current direction for the mesh I imagine "walking around the loop" in the direction of the current, and record the rises and falls of the potential as I pass through components.

Consider: start at the "-" terminal of the 10V supply and moving through that supply in the direction of the current the potential *rises* by 10V. That is, +10V. Proceeding on to move through R1 the potential will *drop* by i₁*R1, and so on. So:

10V - i₁*R1 - i₁*R2 + 12V = 0.

That's how I do it, but your equation will yield the same result for the current.

 

[NascentOxygen]

[bI'm kind of confused as to how to approach this question as a whole because I'm getting put off from the two voltage sources.

You could analyze it using superposition. Having recognized that i₂ is zero, the analysis is much simpler.

First, pretend source S₂ is replaced by a short circuit (0 volts and 0 ohms), and the circuit can be see to be a potential divider fed by S₁. Determine V_ab.

Next, pretend S₁ is replaced by a short circuit (a solid wire). You again have a potential divider, but this time it is fed by S₂. Determine V_ab in this arrangement.

Now, because the circuit is composed of linear elements, the actual value of V_ab will be the sum of its two components determined above.

 

[ohdrayray]

Thanks so much for the help :) but I just wanted to make sure that my answer was correct so:

-10 + 1200*i₁ + 3300*(i₁ - i₂) - 12 = 0
since i₂ = 0 A,
-22 + 4500*i₁ = 0
i₁ = \frac{22}{4500} A = 0.0049 A

Then, for V_oc:
3300*(i₂ - i₁) + V_oc + 12 = 0
since i₂ = 0 A,
3300*(\frac{-22}{4500}) + V_oc + 12 = 0
V_oc = \frac{62}{15} V = 4.13 V