Problem 55-3
Source: Problem elaborated by the author of the site.
Calculate the current I and the average power dissipated by the
circuit show in the Figure 55-3.1.
Figure 55-3.1
Solution of the Problem 55-3
In the figure above, note that the resistor of 4 ohms is in
parallel with the capacitor
-j12, both connected to point b. Therefore, we can calculate the
impedance of this parallel, which will be called Zb.
Zb = 4 XC / (4 + XC ) = 4 (-j12) / (4 - j12 )
Performing the calculate:
Zb = 3.60 - j1.20 = 3.80 ∠-18.43°
Note that Zb is in series with the resistor of 6.4 ohms and the inductor of
j1.20. Calculating this series (called Zs):
Zs = 3.60 - j1.20 + 6.40 + j1.20 = 10 ohms
This impedance, in turn, is in parallel with the capacitor of - j30, as can
be seen in the Figure 55-3.2.
Figure 55-3.2
Calculating this parallel determines the impedance at point a, called Za, or:
Za = 10 XC / 10 + XC = 10 x (-j30) / 10 - j30
Performing the calculation:
Za = 9 - j3 = 9.49 ∠-18.43°
And, finally, adding this impedance to the value of the inductor j9 (which is in series)
we can calculate the equivalent impedance that the circuit offers to the source of tension. Soon:
Ztotal = 9 - j3 + j9 = 9 + j6 = 10.82 ∠33.69°
With all these data we are able to calculate the current I by applying the
Ohm's law, or:
I = V / Ztotal = 30 ∠+30° / 10.82 ∠33.69°
Performing the calculation:
I = 2.77 ∠-3.69°
Having the value of I and V, we can calculate the apparent power of the circuit, or:
S =V I* = 30∠30° x 2.77∠3.69° = 83.1∠33.69° VA
With the value of S we can calculate the average power and the reactive of the circuit. So:
P = 83.1 x cos (33.69°) = 69.14 W
Q = 83.1 x sen (33.69°) = 46.1 VAr
Do not forget that the average power is the power dissipated only by resistors. In this case, the resistors of 6.40 and 4 ohms. Inductors and capacitors do not fit into this account.
