Homework Help: Circuit Analysis Question.

1. Oct 30, 2013

eyeanand

1. The problem statement, all variables and given/known data
The problem statement is in the Attached Image...To find the Voltage "V"

2. Relevant equations

I tried aplying Thevenin theorem here...so Calculated Rth looking in to the circuit with Voltage shorted and current sources opened.

3. The attempt at a solution

by inspection i get Rth = infinity and and Isc = 2A ....hence Vth = infinity x 2A = infinity.

I do not get whether the answer or the approach is correct hence kindly help.

Attached Files:

• Circuit.jpg
File size:
13.2 KB
Views:
104
2. Oct 30, 2013

Staff: Mentor

It's an odd circuit, no doubt. The series-connected ideal 2A current source trumps all the rest of the component behaviors as far as what the load sees is concerned.

An ideal current source will produce any potential difference necessary to maintain its defined current. So the 5V voltage source and 2 Ω series resistance will be nullified (or rather, compensated for) by the current source demanding that the load accept 2A no matter what. The take-away message is: no matter what you put in series with an ideal current source, the whole lot still looks like the same ideal current source.

So, what must the potential V be in order that the load (no matter what load) pass 2A? I think you'll have to answer with an expression rather than a value.

3. Oct 31, 2013

CWatters

I agree. You cannot determine V from the problem. Perhaps the answer is "V is always indeterminable without knowing the voltage across the current source"

If Vc is the voltage across the current source then all of the following appear valid...

V Vc
4 5
5 4
-1 +10
-1000 +1009
+909 -900