Problem 52-13
Source: Problem prepared by the author of the site.
Knowing that the power dissipated by the circuit shown in Figure 52-13.1 is 960 W with
PF = 0.6 capacitive and that 4 R = 3 X, determine the values of R, X and XC.
Figura 52-13.1
Solution of the Problem 52-13
Let's take the current from the current source of 10 A as a reference. So let's write I = 10∠ 0°.
So, if the circuit dissipates P = 960 W with power factor 0.6, it is possible to calculate the apparent power S of the circuit, or:
S = P / FP = 960 / 0.6 = 1,600 VA
Knowing the apparent power and the real power, then we can calculate the reactive power of the circuit.
Q = √(S2 - P2) = √(1,6002 - 9602) = - 1,280 VAR
Note the minus sign in the Q value, as the circuit is capacitive. Now we can write the apparent power in its form
cartesian.
S = P + j Q = 960 - j 1,280 VA
Therefore, knowing the values of I and S, we can find the voltage value over the current source, or
Vbd.
Vbd = S / I = (960 - j 1,280) / 10
Carrying out the calculation, we get:
Vbd = 96 - j 128 = 160 ∠- 53.13° V
Knowing the value of Vbd we can easily calculate the value of I1 using the
Ohm's law, that is:
I1 = Vbd / - j20 = (96 - j 128) / - j20
Carrying out the calculation, we get:
I1 = 6.4 + j 4.8 = 8 ∠36.87° A
From the circuit shown in Figure 52-13.1, we can easily see that I1 + I2 = 10. Like
we know the value of I1 so we can calculate the value of I2.
I2 = 10 - I1 = 10 - (6.4 + j 4.8)
Carrying out the calculation, we get:
I2 = 3.6 - j 4.8 = 6 ∠- 53.13° A
Knowing the value of I2 it is possible to calculate the voltage Vcd.
Vcd = R1 . I2 = 10 (3.6 - j 4.8)
Carrying out the calculation, we get:
Vcd = 36 - j 48 = 60 ∠-53.13° V
At this moment it is possible to calculate Vbc. So:
Vbc = Vbd - Vcd = 96 - j 128 - (36 - j 48)
Carrying out the calculation, we get:
Vbc = 60 - j 80 = 100∠- 53.13° V
On the other hand, let's calculate the power dissipated by the resistor R1.
PR1 = R1 . |I2|2 = 10 x 62 = 360 W
With the calculation of this value we can calculate the power dissipated by the resistor R, because P = PR +
PR1. Soon:
PR = P - PR1 = 960 - 360 = 600 W
With the given 4 R = 3 X we can write the impedance Z as Z = R + j X = R + j 4/3 R and , consequently,
the modulus of impedance Z is given by |Z| = 5/3 R. In this way, we can calculate the value of the current
IR.
|IR| = |Vbc| / |Z| = 100 / 5/3 R = 60 / R
As we know the value of PR, and knowing that:
PR = R |IR|2 ⇒ 600 = R x (60/R)2
Solving this easily solved equation we find the value of R, or:
R = 6 Ω
And considering the relation 4 R = 3 X we easily find the value of X, or:
X = 8 Ω
Therefore, we can write the impedance Z as:
Z = 6 + j 8 = 10∠53.13° Ω
And, finally, to find the value of XC we must calculate the reactive powers. We know that
Q = - 1,280 VAR and that |IR| = 10. Let's calculate QX.
QX = X |IR|2 = 8 x 102 = 800 VAR
And the reactive power for capacitor C1 is given by:
QC1 = - |XC1| |I1|2 = - 20 x 82
= - 1,280 VAR
With the calculation of these values it is possible to calculate the value of QXc,
because Q = QX + QXc + QC1. Soon:
