# Basic AC circuit analysis w/ wattmeter

I am having trouble dealing with this exercise. The data given is: $U_G=110V, \: \omega=150rad/s, \: V_1=0V, \: V_2=5V, \: W=650W$ Power factor =1.

Being the power factor equal to 1, I should be able to get the current intensity using the active power formula right? $P= U_G \cdot I \cdot cos \varphi$ so $\frac{650}{110}=I$. As the total intensity is in phase with the voltage.

However I don't know how to proceed from there in order to find: $R_1 , R_2, C,L$

• Well you have $V_2$ across $R_2$ so work out the current through $R_2$ and you have the current through C. You also know the voltage across L ... Mar 20 '18 at 18:08

simulate this circuit – Schematic created using CircuitLab

Figure 1. Since V1 (on original diagram) = 0 then all these circuits are equivalent.

Since you know the voltage across R2 you should be able to work out the rest.

• Excuse me, but your right bottom version should be removed. To see it have only pure resistors R1=1 Ohm, L1=2 Ohm, C1=1 kOhm, R2=2 kOhm. and compare total resistances Mar 20 '18 at 21:25
• I do believe that you are correct and that I went a step to far in my simplification. Fixed, thank you. Mar 20 '18 at 22:07
• thanks! so, I could get the R1+R2 value, by using P=V^2/R and then KVL for solving the 4 unknowns, right? I feel like I am missing an equation for solving the 4-unknowns system. Could we also assume that the equivalent impedance will have no imaginary part because the PF=1? And thus forcing the imaginary part equal to zero, could we get the missing equation. Mar 21 '18 at 23:17

The big clue is that V1 = 0 volts and this tells you (if you do a little maths) that: -

$\dfrac{L}{R_1} = CR_2$.

V1 being zero also tells you that the voltage across R1 plus the voltage across R2 equals $U_g$. And V2 is the voltage across R2 hence from this you can find R1 and R2 (because you know the power and that power is only dissipated in the resistors).

Can you take it from here?

Having said the above, I believe that V2 should be 55 volts and not 5 volts. This is because if there is unity PF, the current through C has to be the same magnitude as the current through L AND their phase angles should cancel to zero.