Problem 55-5
Source: Problem created by the author of the site.
For the circuit shown in the Figure 55-5.1, calculate:
a) the current I.
b) the voltage Va.
c) the average power supplied to the circuit.
Figure 55-5.1
Solution of the Problem 55-5
Item a
For a better understanding of the circuit, in the Figure 55-5.2, the topology of
the circuit was changed, in order to clarify that the branch where the series has
formed by the resistor of 15 ohms with the inductor of j15, is
actually in parallel with the voltage source. Calling this circuit of
Zs, is clear by the figure that circulates the current
I2 by it. Like this:
Zs = 15 + j15 = 15 √2 ∠+45°
Calculating the current I2, we have:
I2 = V / Zs = 100 ∠40° / 15 √2 ∠+45°
Performing the calculation:
I2 = 4.71 ∠-5° = 4.70 - j0.41 A
Figure 55-5.2
In the point a, we have the parallel of the capacitor of - j15 with the resistor of
20 ohms. Then the impedance, calling it Za, will be:
Za = 20 XC / 20 + XC = 20 x (-j15) / 20 - j15
Performing the calculation:
Za = 7.20 - j9.60 = 12 ∠-53.13°
Adding Za to the 10 ohms resistor (which is in series
with Za), we obtain the impedance of this circuit, which will be called Zfon. Then:
Zfon = 17.20 - j9.60 = 19.70 ∠-29.17°
Then, the current I1 will be:
I1 = V / Zfon = 100 ∠ 40° / 19.70 ∠-29.17°
Performing the calculation:
I1 = 5.08 ∠+69.17° = 1.81 + j4.75 A
But, as the current I = I1 + I2, then:
I = 6.51 + j4.34 = 7.82 ∠+33.69° A
Notice that this angle is less than the angle of the voltage V, meaning that the circuit has inductive predominance, since the current is delayed with respect to the voltage V.
Item b
As we know the value of Za and of I1,
by applying the Ohm's law, we get:
Va = Za I1 = 12 ∠-53.13° x 5.08 ∠+69.17°
Performing the calculation:
Va = 60.96 ∠+16.04° V
Item c
To calculate the average power, we must
find the apparent power.
S = |V| |I| = 100 x 7.82 = 782 VA
We need to find the angle of lag between the voltage V and the current
I. Since the angle of the voltage is + 40° and that of the current 33.69°,
then the value of φ is equal to φ = 40 - 33.69 = 6.31°. Therefore, using the equation of
average power and performing the calculation:
P = S cos φ = 782 x cos 6.31° = 777.26 W
Notice that this angle of 6.31° is the angle that should be used if the problem calls for the
power factor value of the circuit. Like this:
FP = cos φ = cos 6.31° = 0.994
Of course, the power factor is inductive. However, the power factor is very close to ONE, indicating that although the circuit contains reactive elements, it behaves "almost" as a purely resistive circuit. The average power calculated as above, is totally dissipated as heat (Joule effect) by the three circuit resistors.
Complementation
To complement the problem solution, the graph in the Figure 55-5.3 shows
the currents and the voltage V. Note that the phasor sum of I1 and I2 results in the current I.
It also shows the delayed current of 6.31° = 40° - 33.69° with respect to the voltage V.
