0
\$\begingroup\$

I am asked to find the open circuit voltage v(AB).

So far, I have found the current, I(1) by using the current divider, and I got approximately: 0.307A.

Except, I am not sure where to go on from here to find the open circuit voltage.

Any help would be appreciated. Thank you very much. enter image description here

\$\endgroup\$
2
  • 1
    \$\begingroup\$ If you know the current through the 68 ohm resistor, don't you know the voltage across it? \$\endgroup\$ May 15, 2018 at 12:18
  • \$\begingroup\$ Thank you so much. I did not even consider what I had calculated using the voltage divider. Very silly question, thank you for your time. \$\endgroup\$
    – Angela
    May 15, 2018 at 12:20

3 Answers 3

2
\$\begingroup\$

Open circuit voltage is the voltage which is across 68 ohm resistor. Open circuit voltage is always the voltage of the component to whom the open circuit is parallel.

\$\endgroup\$
1
  • \$\begingroup\$ That works in this case, but sometimes there is not a convenient single component with which the open circuit voltage is in parallel. In the more general case, the open circuit voltage is simply the voltage when there is no load. \$\endgroup\$ May 16, 2018 at 10:35
2
\$\begingroup\$

All you do is apply ohm's law to solve this.

first find the source voltage by finding the load resistance and multiply that by the known current.

so \$ \frac1{ \ \frac {1}{47}+\frac {1}{22+68} } \$= \$ \frac1{ \ \frac {1}{47}+\frac {1}{90} } = 1/(0.0213 + 0.0111)= 1/0.0324= 30.876 \ \Omega \$

0.750 A * 30.876 ohms= 23.157 V

find the current in the 22 ohm and 68 ohm branch.... \$\frac{23.157\ V}{90\ \Omega} = 0.257\ A\$

now lets see what the voltage drop across the 68 ohm resistor....

0.257299269 * 68= 17.496350292 V

\$\endgroup\$
13
  • 1
    \$\begingroup\$ -1 and voting to delete. Not only does this attempt to directly answer a homework question, but is absurdly deep into false precision, and it's wrong in addition. \$\endgroup\$ May 15, 2018 at 18:35
  • \$\begingroup\$ 1. I don't know that this is a "Homework Question" @OlinLathrop 2. I guess I could round if that will make you happy. 3. Ohm's law hasn't changed, neither has kirchoff's law and its quite accurate. \$\endgroup\$
    – drtechno
    May 15, 2018 at 19:05
  • \$\begingroup\$ but I will round it if you want. \$\endgroup\$
    – drtechno
    May 15, 2018 at 19:05
  • 1
    \$\begingroup\$ MathJAX fraction syntax is \$ \frac {top}{bottom} \$ which gives \$ \frac {top}{bottom} \$. See MathJAX basic turorial. \$\endgroup\$
    – Transistor
    May 15, 2018 at 20:49
  • 1
    \$\begingroup\$ @awj: No, this answer is also wrong. Note the polarity of the current source and how Vab is defined. Maybe I should have also mentioned the incorrect units earlier. The last line, for example, shows the product of two dimensionless numbers resulting in a value of Ohms. Clearly that can't be right. \$\endgroup\$ May 15, 2018 at 21:51
1
\$\begingroup\$

One way to solve this is to progressively simplify the circuit into either a single Thevenin or Norton source.

For example, if going the Thevenin route, combine the current source and the 47 Ω resistor to make the equivalent Thevenin source. Now combine that with the 22 Ω resistor to make the equivalent Thevenin source of everything left of the 68 Ω resistor. Now you have a simple problem of a Thevenin source with a resistive load. It is now simple to solve for the voltage across the 68 Ω resistor.

\$\endgroup\$
2
  • \$\begingroup\$ Would whoever downvoted this, would you care to explain what is wrong, or is this just a revenge downvote because of my downvote on the attempt to answer a homework question? \$\endgroup\$ May 15, 2018 at 21:49
  • \$\begingroup\$ This is probably the best way to solve this, and it shows the steps to get to the solution too, without directly answering the homework question. So the downvote is incorrect. \$\endgroup\$
    – MCG
    May 18, 2018 at 11:56

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.