Problem 55-14
Source: Problem elaborated by the author of the site.
From the phasor diagram shown in the Figure 55-14.1 determine the equivalent circuit and the angle θ,
knowing that |V| = 200 V, |I1| = 10 A e |I2| = 8 A.
Solution of the Problem 54-14
Item a
From the figure, note that the voltage and currents are referenced to the axis x.
Thus, we can write the phasor values of these variables by taking the axis x as a reference,
that is, V = 200 ∠-30° , I1 = 10 ∠30° and
I2 = 8 ∠-60°. In addition I1
is advanced by 60° in relation to Vand this characterizes a circuit with
capacitive predominance. As we know V and
I1, then it is possible to calculate this impedance, denoting it by
ZC and substituting the numerical values, in the equation below, we find:
ZC = V / I1 = 20 ∠-60° = 10 - j 17.32 Ω
On the other hand, I2 is delayed at 60° - 30° = 30° with respect to V, characterizing an impedance with inductive predominance. Denoting this impedance by
ZL and after the substitution by the numerical values, in the equation below, we obtain:
ZL = V / I2 = 25 ∠30° = 21.65 + j 12.5 Ω
We must calculate the value of the modulus of IT.
As I1 and I2 make an angle of 90° between them, we can use
Pythagoras. Like this:
IT = √ (I12 + I22) = √(102 + 82) = 12.8 A
In order to calculate the angle θ, which is the angle between IT
and the axis x, first calculate the angle between IT and I1
calling it φ. Thus, having the value of this angle and subtracting from 60° will find the value of
θ.
With the value of IT , I1 and I2, we can use the law of cosines to find the value of φ, using the eq. 51-01. After the calculation, we find:
φ = 51.34°
And the value of θ is:
θ = φ -60° ⇒ θ = -8.66°
With the values of ZC and ZL calculated above, one can
draw the electrical diagram of the circuit. There are several circuits that meet this specification,
however, a very simplified circuit is shown in Figure 55-14.2.
Item b
We have already calculated the module of IT and θ. Writing in phasorial form:
IT = 12.8 ∠-8.66° A
Knowing the value of V and IT, we can calculate the total equivalent impedance that the circuit offers to the voltage source. Like this:
Zeq = V / IT = 200∠-30° / 12.8∠-8.66
Performing the calculation:
Zeq = 15.625∠-21.34° = 14.55 - j5.69 Ω
Note that the equivalent impedance has capacitive predominance. This conclusion can also be reached by looking at the phasor diagram and verifying that IT is advanced of 30° - 8.66° = 21.34° in relation to V. Of course you can get the value of Zeq calculating the parallel between ZC and ZL.
The result has to be the same. Check!!!!.