Open Voltage and Short Circuit

[ohdrayray]

Homework Statement

So the first part is to find Voc in this circuit:

So I did this:
-10 + 1200*i₁ + 3300*(i₁ - i₂) - 12 = 0
since i₂ = 0 A,
-22 + 4500*i₁ = 0
i₁ = 22/4500 A = 0.0049 A

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

Then the next part is to find the short circuit current Isc in the same circuit:

Assuming that my calculations for i1 were right, let i1 = 0.0049
Isc = i2 so then do mesh current analysis for the second mesh, which will get:

3300(i₂ - i₁) + 12 = 0
3300(i₂ - 0.0049) + 12 = 0
3300i₂ - 16.137 + 12 = 0
3300i₂ = 4.137
i₂ = 0.001254 A
.:. I_sc = 0.00125 A

Would this be correct? Thanks in advance for any help given. :)

 

[gneill]

Homework Statement

So the first part is to find Voc in this circuit:

So I did this:
-10 + 1200*i₁ + 3300*(i₁ - i₂) - 12 = 0
since i₂ = 0 A,
-22 + 4500*i₁ = 0
i₁ = 22/4500 A = 0.0049 A

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

Since you know that terminals a and b are open circuited, you only have one loop. Writing in i₂ as a variable is unnecessary.

You've solved for the current in the loop, and the result is okay. You might consider taking advantage of unit prefixes. Thus i₁ = 4.9 mA.

Now, regarding V_oc, note the polarity of the 12V supply: starting at terminal b and proceeding through the 12V supply there is a DROP of 12V. Proceeding on through R2, since the current i₁ flows from top to bottom of R2, there will be a potential rise as we pass through. So V_oc = -12V + i₁*R2.

Then the next part is to find the short circuit current Isc in the same circuit:

Assuming that my calculations for i1 were right, let i1 = 0.0049

Nope! You can't do that since the circuit changes when the open circuit at a-b becomes a short circuit; You now have two loops instead of one, and they will interact via their common components (R2 in particular).

You'll have to write the two KVL mesh equations and solve for the currents, or use another method as you see fit (Source transformations (Norton), superposition, etc.). Superposition would be particularly simple for this problem.