Engineering Is My Open Circuit Voltage and Short Circuit Current Calculation Correct?

AI Thread Summary
The calculations for open circuit voltage (Voc) and short circuit current (Isc) in the circuit were initially approached correctly, but some assumptions were flawed. The first part correctly calculated Voc as 4.13 V, but the method could be simplified by recognizing that only one loop exists when terminals are open. For Isc, the assumption that i1 remains the same when the circuit changes to a short circuit is incorrect; two loops need to be analyzed due to the interaction of components. It is recommended to use mesh analysis or superposition to accurately determine Isc. Overall, while the initial calculations were on the right track, adjustments are necessary for accurate results.
ohdrayray
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Homework Statement



So the first part is to find Voc in this circuit:
263bz3c.png


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

Then the next part is to find the short circuit current Isc in the same circuit:
2uona04.png


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
.:. Isc = 0.00125 A

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

Homework Statement



So the first part is to find Voc in this circuit:
263bz3c.png


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 Voc:
3300*(i2 - i1) + Voc + 12 = 0
since i2 = 0 A,
3300*(−22/4500) + Voc + 12 = 0
Voc = 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 Voc, 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 Voc = -12V + i1*R2.

Then the next part is to find the short circuit current Isc in the same circuit:
2uona04.png


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.
 

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