The problem is....Find the Thevenin equivalent looking into terminals a-b of the circuit and the current ix. V=60V and I=7A.

If u look at my work, i got Vth and Rth correct,( i checked) i am just not able to get the value for Ix correct.

The Problem

My Work

  • \$\begingroup\$ I don't think your analysis is correct. For V_TH you should take out the 6ohm resistor (open circuit it) off the circuit and look at the node ab. The difference between Vb and Va is V_TH. For R_TH you must short circuit the voltage source and open circuit the current source and then measure the equivalent resistance between node a and b. Do it again. \$\endgroup\$
    – dirac16
    Commented Feb 4, 2017 at 21:12
  • \$\begingroup\$ that's what i had done originally, but then the professor sent an email saying that the 6ohm resistor is actually part of the circuit. Also when i inputed 3.75 ohms and -15/8 V it was correct. \$\endgroup\$
    – J. Doe
    Commented Feb 4, 2017 at 21:17
  • \$\begingroup\$ Yes, it's part of the circuit and you'd have to take it back in place after you are done with V-Th and R_Th. Did you do that? \$\endgroup\$
    – dirac16
    Commented Feb 4, 2017 at 21:24
  • \$\begingroup\$ i believe that i did. \$\endgroup\$
    – J. Doe
    Commented Feb 4, 2017 at 21:52
  • \$\begingroup\$ I did some quick calculations and got the same V-Th as yours. @dirac16 I'm not sure if I understood you correct but, if you take out the 6 Ohm resistance your just end up calculating the equivalent for the same circuit just without the 6 Ohms, so I don't rly so see how this is supposed to be helpful? \$\endgroup\$ Commented Feb 4, 2017 at 22:27

2 Answers 2


The thevenin equivalent voltage shows the open clamp voltage between A and B. During the calculations your resistor was already connected, you shouldn't use the equivalent resistance, but simply the resistor between A and B. You just want to know the current trough a resistor with a known voltage over it.

$$I = \frac{V}{R} \text{ where } R = 6Ω$$

Asuming the 6Ω resistor is part of the circuit your other calculations seem to be correct indeed.

  • \$\begingroup\$ That's what i had originally thought, but when i inputed 6 as the answer, it came back incorrect. \$\endgroup\$
    – J. Doe
    Commented Feb 4, 2017 at 21:02
  • \$\begingroup\$ Uh, the current is not 6... You should calculate the current using the voltage and the 6 ohm resistor. \$\endgroup\$
    – Douwe66
    Commented Feb 4, 2017 at 22:16

a few simple steps:

1) convert the voltage source + 10R serial resistor into a current source // 10R parallel resistor;

2) parallel the two resistors into a 5ohm resistor;

3) convert that current resistor // 5ohm parallel resistor into a voltage resistor + 5R serial resistor;

4) convert the current source on the right + its parallel resistor into a voltage source + 10R serial resistor.

5) the rest is easy.


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