0
\$\begingroup\$

I'm currently trying to gain an understanding of Thévenin and Norton equivalent circuits and I've run into a circuit I simply don't understand. I've been researching online, but I've just run into contradictory information that often isn't helpful to begin with.

Here's the circuit: Here's the circuit:

I could simplify this circuit, but that ideal current source is my big roadblock. I'm currently trying to find the voltage between A and B.

I don't even know where to start. If someone could show me the right direction, that would be great! Thanks

\$\endgroup\$
  • 3
    \$\begingroup\$ If you don't see any clever ways to simplify the circuit, you can just resort to superposition. \$\endgroup\$ – The Photon Oct 4 '16 at 4:23
  • 4
    \$\begingroup\$ The voltage betwen A and B is simply 3A x 48 Ohm \$\endgroup\$ – Claudio Avi Chami Oct 4 '16 at 7:35
  • \$\begingroup\$ If you replace the current source with open circuit (because its internal impedance is infinity, theoretically) then you'll see that the voltage across 48Ohm resistor is not affected by 5V voltage source. Thus, as @ClaudioAviChami stated, voltage across 48Ohm resistor is simply 3A x 48Ohm. \$\endgroup\$ – Rohat Kılıç Oct 4 '16 at 19:46
1
\$\begingroup\$

The leftmost 100 ohm resistor from your schematic can be excluded form analysis - voltage across it is constant at 5V. Let assume that current in the 160 ohm resistor is I2 and current in 50+50 ohm resistor is I1. One than can write following equations : I1 + I2 = 3A, I2*160 - 5V - I1*100 = 0. Hope this helps.

schematic

simulate this circuit – Schematic created using CircuitLab

|improve this answer|||||
\$\endgroup\$
0
\$\begingroup\$

There is no mystery here. The Vab is shown below,

enter image description here

The software allows you to double check the result, however, to determine the actual voltage in Thévenin configuration, you turn on the independent sources and determine the voltage at AB. (Hint: see Ohm's law).

|improve this answer|||||
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.