0
\$\begingroup\$

In the circuit below the complex voltage \$u\$ should be calculated using Kirchoff's law. I know this law implies that sum of voltage should be zero. I'm trying to solve it like this but I am not sure if I aam doing correct steps:

enter image description here

UPDATE

enter image description here

\$\endgroup\$
4
\$\begingroup\$

Sean, I'm afraid you need to review some basic circuit theory. The first and second pages of work reveal some serious misconceptions.

It looks like I1 and I2 are labelling mesh currents while I3 labels a branch current and then you're writing an equation involving them. When you try to do that, all you get is gibberish. Mesh current analysis involves writing KVL equations, not KCL equations.

stevenh gave a very good hint at how to proceed using node voltage analysis. [UPDATE: I see that while I was typing this, stevenh updated his answer so the rest of my answer more or less duplicates his added info.] In node voltage analysis, you pick one node to be your zero node or "ground" node. Think of it as the node you place the black lead of your voltmeter on. stevenh's suggestion to use the node that the lower end of the resistor connects to is an excellent one.

Then, you write KCL equations for each remaining unknown node voltage (where the red lead of the voltmeter goes).

In this case, there's only one unknown node voltage and it is, in fact, the voltage you seek, \$V_1\$. The voltage across the resistor is in fact the node voltage at the node the top of the resistor connects to. So, you write a KCL equation for that node.

To do KCL, you set the sum of the currents entering (or exiting) the node equal to zero.

The currents exiting the node are the current to the left through the current source, the current down through the resistor, and the current to the right through the inductor. The sum of these currents is:

\$j3A + \dfrac{V_1}{4 \Omega} + \dfrac{V_1 - (-j6V)}{j4 \Omega} = 0\$

Solve for \$V_1\$ and you're done.

By the way, you can write the answer by inspection using superposition. The first term involves the current source and current division. The second term involves the voltage source and voltage division.

\$\endgroup\$
4
\$\begingroup\$

No, that's not correct. You say that the voltage across the resistor is 12j V, but that assumes that all of the 3j A go through the resistor, and there's also the path through the voltage source and inductor.

I would call the lower connection of the resistor ground, and the upper V1. Now try to find V1 by applying KCL to the node (that's the sum of three currents).

edit (re your update of the question)
Your arrows suggest that you want to want to find the currents in the two loops, but that's not how KCL works. This doesn't cause anything but problems: I1 seems to be -I2 if I look at how they flow through the resistor, and I3 = I2.

KCL: you pick a node: V1. Then assign a current name to all branches that start from V1: I1 is the current source at the left, I2 is the current from V1 through the resistor to ground, and I3 is the current from V1 through the inductor and the voltage source to ground. Then I1 + I2 + I3 = 0. I1 is easy, that's given: 3j A. I2, also easy, a voltage difference over a resistor, write the equation for I2. For I3 you have two voltage sources, V1 and the -6j V source, and one impedance: the inductor. Again, write down I3 as a difference between two voltage sources divided by an impedance.

Now you have three expressions for the currents, set their sum equal to zero, and you have an equation with just one variable: V1. Calculate V1, and fill its value in the equations for I2 and I3, and you're done.

\$\endgroup\$
4
  • \$\begingroup\$ Thanks. You mean replacing resistor by open? \$\endgroup\$
    – Sean87
    Jul 3 '12 at 15:21
  • \$\begingroup\$ No, leave everything as it is. You have three current paths from V1 to ground, describe them as function of V1. Then use the equality I1 + I2 + I3 = 0 (mind the direction of the currents) to find V1. If you know V1 you can fill that value in the equations for the currents. \$\endgroup\$
    – stevenvh
    Jul 3 '12 at 15:27
  • \$\begingroup\$ I'v updated the question with what I underestood can you check again please? \$\endgroup\$
    – Sean87
    Jul 3 '12 at 15:48
  • \$\begingroup\$ Updated my answer. \$\endgroup\$
    – stevenvh
    Jul 3 '12 at 16:02

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.