Yes, albeit, strictly speaking, 'phase' is not properly a property of reactance itself.... Anyway - good work!right?
That is exactly correct. Notice that the voltage source Vs is only supplying 10 volts, but you have 60 volts appearing across the inductor. This is what a series resonant circuit can do for us. It is an important phenomenon that we can use for impedance matching.OK, As the circuit is at resonance mean both Xc and Xl are out of 180 phase cancel each other.
the total current wil be E=I*R
I=10/5=2Ams
so, Exl=2*30=60V i think
No. You have R1 and C1 in series, with that combination in parallel with L1, like this:The total impedance by shoring the Voltage source and open node 3
will give us:
Z=√(R²+Xc²)
Z=30.41ohm, right?
Ok. I see your problem here, RRITESH. You are having difficulty working with complex impedances.OK,
At 30.41ohm // with L1 30ohm
Z=1/(30.41²+30²)
Z=42.7ohm may be.
OK. But carry the multiplication in the numerator further, so you end up with an expression in the form A + j B.(5-30j) // (0+30j)
(5-30j)*(0+30j)/ (5-30j)+(0+30j)
(5-30j)*(0+30j)/(5)
...........
..............
It's simple complex arithmetic. I think you must have forgotten how to do this.How to do?
(5 - j30) * (0 + j30) =
5*(0 + j30) - j30*(0 + j30) =
5*0 + 5*j30 - j30*0 - j30*j30 =
0 + j150 - 0 - j*j*900 =
In this next step it's very important to remember that j*j = -1
0 + j150 - 0 - (-1)*900 =
900 + j150
Well, actually reactive power has a use. It performs the impedance transformation we want to happen in this case, but in the end we want to get rid of it.Now i got it that because the reactive power and impedance has no use.
Why we say it is imaginary??