Problem 55-12
Source: Problem elaborated by the author of the site.
In the circuit shown in the Figure 55-12.1, calculate the value of Z to meet the conditions imposed by the problem data.
Data: V = 10 cos (100t + 15°) and i = 0.1 sen (100t + 47.6°).
Figure 55-12.1
Solution of the Problem 55-12
Since the value of V was given as a cosine function and the current as a function
sine, the functions must be made compatible. By transforming the voltage V, writing it as a
sine function, we get:
V = 10 sen (100t - 75°)
From the above equation, we realize that the value of ω (term that accompanies the variable
t) is:
ω = 100 rad/s
Thus, we can calculate the values of the inductive and capacitive reactances, that is:
jXL1 = jω L1 = 100 x 0.01 = j Ω
jXL2 = jω L2 = 100 x 0.22 = j22 Ω
-jXC = -j / ω C = -j / (100 x 0.05 ) = -j0.2 Ω
Let's define two impedances between the points a-b: ZS as the impedance of the upper branch and ZI as the impedance of the lower branch. Soon:
ZS = jXL1 + j XC = j1 - j0.2 = j0.8 = 0.8 ∠+90°
ZI = R + j XL2 = 100 + j 22 = 102.4 ∠12.4°
The impedance of this parallel circuit will be:
ZP = ZS ZI / ZS +
ZI = 81.92∠102.4° / 102.57∠12.84°
Then, performing the calculation:
ZP = 0.8 ∠89.56° ≈ j 0.8
To find the solution to the problem, we must calculate the total impedance that the circuit offers to the voltage source. For this, the voltage Vab must be calculated
and then the current i2. Then:
Vab = ZIi = 102.4∠12.4° x 0.1∠47.6°
Performing the calculation:
Vab = 10.24∠60°
In possession of the value of Vab, easily calculates the value of
i2:
i2 = Vab / ZS
= 10.24∠60°/ 0.8∠90° = 12.8∠-30°
It is known that I =i + i2:
I = 0.1∠47.6° + 12∠-30° = 12.022∠-29.54°
Calculating the value of the total impedance, ZT:
ZT = V / I = 10∠-75° / 12.022∠-29.54°
Performing the calculation:
ZT = 0.83∠-45.46° = 0.582 - j 0.592
On the other hand, it is known that the total impedance ZT is equal to the sum of the impedances of the circuit, that is:
