[Math] Solving circuit analysis problems

[nForce]

Hi,

this topic will be a little refresh of math and circuit analysis. The first circuit is here:



Here U = Voltage.
How do we get U_1? I know that here is used a voltage divider, but I can't see it.

More problems to come 

[Andy Watson]

I get the feeling that there is more to the original question that you are telling us.

With the given diagram, my first step would be to write-down the impedance for the parallel combination of \$L\$, and \$R + Z_b\$.

[nForce]

I forgot to mention that L and R are inside two-port network. Those circles are contacts. Does this help you?

[Andy Watson]

I forgot to mention that L and R are inside two-port network. Those circles are contacts. Does this help you?

Only to confirm my suspicions that there is more to this question than you have shown

You asked "how do we get \$U_1\$. Since an equation for \$U_1\$ is given I assume you want to know how to get that equation?

As I said, my first step would be to find the impedance of the parallel combination of \$L\$ and \$R + Z_b\$ using the short-cut \$Z_p = \frac{Z_1 Z_2}{Z_1 + Z_2}\$. If you put \$Z_1 = R + Z_b\$ and \$Z_2 = j \omega L\$, I think we have:
$$Z_p = \frac{(R + Z_b) j \omega L}{R + Z_b + j \omega L}$$
Now you have your divider of \$Z_g\$ and \$Z_p\$ - plug those into the first equation given and the equation for \$U_1\$ readily drops out.

However, the diagram is labelled with \$I_1, I_2\$ and \$U_2\$ which leads me to suspect that the original question also specified a method for the solution (e.g. "using Kirchhoff/Thevenin/nodal analysis, "... or whatever), and now you mention "two port network".

[MrAl]

Hi,

I agree with Andy, if you want to see how this works you would first collapse the right side of the network into one impedance Zx, then use that in the voltage divider formula:

U1=Ug*Zx/(Zx+Zg)

Now you are also in a position to calculate the output U2 using a second application of the voltage divider formula:
U2=U1*Zb/(Zb+R)

You dont have to do it that way though.  That's a "reverse collapse".  You can go forward (left to right)  by using Thevenin and Norton source transformations.