U=10 V, R=10 ohm U=10 V, R=10 ohm

I calculated the resistance as such: 1/Rt = 1/R + 1/R <=> Rt = R/2 = 5 ohm

To calculate the voltage I did KVL as such: U - 2Ur = 0 <=> Ur = 5V, where Ur is the voltage in the resistors And then by applying KVL to the closed network that goes from a to b passing through the resistor, I got Ut=5V

Then I calculated the current using U=RI, and I got I=1A. Is this correct? Because in my solutions I=0,5A


I answered your last question here.

Once again, you have the right answer and your "solution" answer is wrong.

You have done the correct method; found the open-circuit terminal voltage, which is the Thevenin voltage. You have then found the Thevenin resistance - also correct. You have then used the transform \$ I_n = V_{th}/R_{th}\$ to find the Norton current, correct again.

Good job - wherever you keep getting your solution answers from, stop using it!!!

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    \$\begingroup\$ It depends where ground reference is; this must be specified. With U=+10 V, as shown, and node b as ground, Vab = Vth = -5 V \$\endgroup\$ – Chu Nov 1 '18 at 7:57
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    \$\begingroup\$ @Chu you are correct, depending on the convention of the way you draw your voltage drop arrows. I've seen it both ways (arrow points to the higher voltage side, or arrow shows which way the current flows). Either way his "solution" be has been supplied is still wrong sadly. \$\endgroup\$ – DSWG Nov 1 '18 at 23:34

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