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U=10V, R=10 ohm What I did was apply kirchoff's voltage law to the closed network with the 2 resistors and the voltage source to calculate the voltage of the resistors. I got U - 2*Uresistor =0 <=> Uresistor =5V

Then, did the same to the closed network with a, b and the one resistor and got that UThevenin=5V

However, the answer is 10V. What did I do wrong?

Also, is there a way to do this by calculating the current first, and then applying the formula U=RI?

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If U is 10V, the answer is definitely not 10V also.

schematic

simulate this circuit – Schematic created using CircuitLab

Two of those resistors form a voltage divider - your answer is correct.

As for your second question - yes, there is a way, but it requires an extra step. The total resistance of the network is R1, in parallel with R2 + R3 in series. This gives the total network resistance as:

\$ R_{eq}= \frac{1}{\frac{1}{R} + \frac{1}{R+R}}\$

We don't need to work that out but - let's just leave it as "Req". The current through the whole circuit is 10V divided by this entire resistance.

\$ I = \frac{V}{R} = \frac{10}{R_{eq}} \$

We now know the total current, but we need the current specifically through R2 and R3 - so we need to do a current divider. Remember, current divider you use the resistance you aren't interested in on top.

\$ I_{R3} = I\times \frac{R1}{R1+R2+R3} = I\times \frac{R}{3R} = I\frac{1}{3} =\frac{10}{3R_{eq}}\$

Now, we know the current through R3. As you have hinted, we can now use V=IR to find the voltage across it - which is the same as the voltage across AB, which is the Thevenin open-circuit voltage we want.

\$ V_{th} =IR=\frac{10}{3R_{eq}}\times R = \frac{10}{3(\frac{1}{\frac{1}{R} + \frac{1}{R+R}})}\times R= \frac{10}{3}\times (\frac{1}{R} + \frac{1}{R+R})\times R = \frac{10}{3}\times (1+0.5) =5V\$

What do you know - we got the same answer! You can see that took a lot longer. The voltage divider law is actually derived from finding the total current, and multiplying by the resistance you want the voltage drop of - that's why it was a lot quicker to just use that.

Finally, if those two weren't enough, here's a simulation screenshot showing that it is definitely 5V. Your answer source is wrong. Good work to you for getting the answer right but :)

enter image description here

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  • \$\begingroup\$ Ok, and what about the thevenin resistance? Because at first I calculated it like this: 1/Rt = 1/2R + 1/R <=> Rt = 2/3R. However I saw a solution that Rt=R/2, and supposedly that was because when you substitute the voltage source with a short circuit, no current is going to flow through the resistor that is in parallel, and as such we eliminate that path from the circuit. The resulting circuit is left with just the 2 resistors, which are now in parallel, and then you do 1/Rt = 1/R + 1/R <=> Rt = R/2. Which is correct? \$\endgroup\$ – chilliefiber Nov 1 '18 at 1:52
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    \$\begingroup\$ The solution is correct, not you (sorry). You have to turn all the sources off, and when a voltage source is turned off it becomes 0V, which is the same as a short-circuit. When it is a short-circuit, R1 is eliminated. The resulting circuit is R2 and R3 in parallel, which gives R/2 Thevenin resistance. \$\endgroup\$ – DSWG Nov 1 '18 at 2:30
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I learned to calculate the Thevenin equivalent voltage and Thevenin equivalent resistance by determining the equivalent open-output voltage ('Thevenin voltage' \$ U_{eq} \$) and the short-circuit output current after which the quotient of the two is the equivalent source's resistance ('Thevenin resistance' \$ R_{eq} \$).

In this case this works out as:

  1. As 'U' is an ideal voltage source, it's impedance is zero and the diagonally placed R doesn't influence the open-output voltage. This voltage is then only determined by the voltage divider \$ R-R \$ and becomes half of U, and as your question implies that \$ U = 10 V \$, it follows that \$ U_{eq} = U/2 = 5 V \$. This is the Thevenin equivalent voltage of the equivalent source.

  2. Then with a shorted load, evaluate the short-circuit output current. This is \$ I_{short-circuit} = 10 V/R \$.

Now the Thevenin equivalent resistance is the quotient of the Thevenin equivalent (open-output) voltage and the short-circuit output current and is \$ R_{eq} = \frac{5 V}{10 V/R} = R/2 \$.

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