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I have following task

enter image description here

I know it's probably really simple (just simple Ohm's law), but I have no clue what kU is. Could anyone explain it to me in the simplest way possible.

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  • \$\begingroup\$ The circle labelled "E" and the one labelled "kU"...are they both supposed to be current sources, or both voltage sources, or is one a source and the other one is an ammeter or what? \$\endgroup\$ – The Photon Dec 13 '16 at 16:09
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    \$\begingroup\$ kU = kV. U is used instead of V for volts in parts of Europe. \$\endgroup\$ – Leon Heller Dec 13 '16 at 16:13
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    \$\begingroup\$ They look to both be current sources but I have never seen that notation used before \$\endgroup\$ – DerStrom8 Dec 13 '16 at 16:14
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    \$\begingroup\$ @LeonHeller While I agree with you in some cases, I don't believe that is the case in this particular circuit. kV doesn't make any sense in this context. \$\endgroup\$ – DerStrom8 Dec 13 '16 at 16:14
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    \$\begingroup\$ It appears the U is the voltage across R, and kU is a voltage dependent voltage source (since the unit of k is V/V) \$\endgroup\$ – Chu Dec 13 '16 at 16:43
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K is a constant that have to be multiplied by U which I guess will be the voltage over R.

The point here is that you don't have to know U value to solve this, I suspect you will be able to get rid of it.

EDIT: Did it myself, you get 2 equation system with 2 unknown parameters U and I. You can solve for both of them.

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  • \$\begingroup\$ You don't have 2 unknown parameters. U depend on I, then you can write 1 equation with only one unknown parameter (from KVL): \$ I= \frac{E-kU}{R} = \frac{E-kRI}{R}= \frac{E}{R}-kI\$. Then \$I=\frac{1}{k+1} · \frac{E}{R} \$ \$\endgroup\$ – Antonio Dec 13 '16 at 16:42
  • \$\begingroup\$ @Antonio Well that is an equation system as I stated above (yes, a simple one) being the second equation U = RI (you just replaced it directly in the second step), since OP wanted "the simplest way possible" didn't wanted to skip any step. Also the thing was that he could solve it by himself without giving him the answer right away. \$\endgroup\$ – xgrimau Dec 13 '16 at 16:50
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If kU is a controlled voltage source, then my hint is you can find the node voltages without even knowing that the third element is a resistor, just use KVL by itself. After you know the node voltages, then you can use Ohm's Law to find the resistor current.

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