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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.
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.
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
\$ \frac {top}{bottom} \$ which gives \$ \frac {top}{bottom} \$. See MathJAX basic turorial.
\$\endgroup\$
May 15, 2018 at 20:49
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.
