Problem 55-17
Source: Problem 10.50 - page 288 -
EDMINISTER, Joseph A. - Book: Circuitos Elétricos (Coleção Schaum) -
2ª edição - McGraw-Hill - 1991.
In the circuit shown in Figure 55-17.1, determine the current I,
so that the voltage between points a-b is Vab = 5 ∠ 30°.
Figura 55-17.1
Solution of the Problem 55-17
Initially we will find the equivalent impedance value between the points a-b which we will call
Zab. Thus, the parallel of j5 with 2 - j2 is:
Zab = (2 - j2) j5 / (2 - j2 + j5)
Performing the calculation, we find:
Zab = (50/13) - j (10/13) = 3.92 ∠ -11.31° Ω
With the value of Zab and how do we know the value of
Vab, we easily calculate the current the value of
I2 according to scheme in Figura 55-17.2.
I2 = Vab / Zab = 5 ∠ 30° / 3.92 ∠ -11.31°
Figura 55-17.2
Performing the calculation, we find:
I2 = 1.28 ∠ 41.31° A
Knowing I2, we can calculate the value of V, that is:
V = ( 10 + Zab ) I2 = 13.87 ∠ -3,18° x 1.28 ∠ 41.31°
Performing the calculation, we find:
V = 17.75 ∠ 38.13° V
At that moment, calculating the value of Z1, as Figura 55-17.2, we can calculate the value of I1 since we know the value of V. So, calculating the parallel of j5 with 2 + j2, we have:
Z1 = (2 + j2) j5 / (2 + j2 + j5)
Performing the calculation, we find:
Z1 = (50/53) + j (90/53) = 1.94 ∠ 60.95° Ω
Therefore, I1 will be:
I1 = V / Z1 = 17.75 ∠ 38.13° / 1.94 ∠ 60.95°
Performing the calculation, we find:
I1 = 9.15 ∠ -22.82° A
In possession of the values of I1 and I2, just perform the
phasor sum of these currents and we find the value of I, or:
