Problem + Hard 55-3
Source: Problem 18 - List of RLC Problems - Discipline
Electrical Circuits of the School of Engineering - UFRGS - 2011 - Prof. doctor Valner Brusamarello.
For the circuit in the figure below, E is in phase with E1 and the
voltmeter V measures 60 volts. Furthermore, X is a component
R , L or pure C. Determine E1, E and X.
Solution of the Problem + Hard 55-3
As the phase of any voltage in the circuit was not indicated, we will take the measured voltage as reference
by the voltmeter, such that V = 60∠0° volts. On the other hand, let's assume that E1 has a phase angle equal to θ. Then I1 will be given by:
I1 = E1∠θ / 7.5 A
The impedance Zcde is calculated so that it is possible to calculate I2.
Zcde = 3 - j 4 = 5∠-53.13° Ω
So I2 will be:
I2 = E1∠θ / 5∠53.13° A
It is known that I is the phasor sum of I1 and I2 , then:
I = (E1∠θ / 7.5) + E1∠θ / (5∠53.13°)
Placing E1∠θ in evidence, we get:
I = ( 1 / 7.5 + 1 / 5∠53.13°) E1∠θ
We know that 1/7.5 = 0.1333 and 1/(5∠53.13°) = 0.12 + j0.16, so adding the two we have :
I = (0.2533 + j0.16) E1∠θ = (0.3∠32.28°) E1∠θ
Note that the voltmeter reading is the phasor sum of the Vac and Vcd voltages.
And of the circuit, Vac = 2 I and Vcd = 3 I2. Then, one can write:
V = 60∠0° = (0.6∠32.28°) E1∠θ + (0.6∠-53.13°) E1∠θ
Now, placing E1∠θ in evidence and carrying out the calculation we have:
(0.882∠-10.45°) E1∠θ = 60∠0°
And then find the value of E1∠θ, or:
E1∠θ = 68∠10.45° V
With the value of E1∠θ, you can calculate the values of E and X .
To do so, let's use the simplified circuit below and relate the variables. Notice that the phase
of E is exactly the same as the phase of E1, as stated in the problem. Also, we add the resistor
of 2 ohms with the element X, through which the current I flows.
Now using the Kirchhoff law for voltage we have:
-E∠10.45° + (2 ± jX) I + 68∠10.45° = 0
Algebraically working the above equation, we get:
((E - 68) / 20.4)∠-32.28° = (2 ± jX)
In the above equation, we used the value of I = 20.4 ∠42.73°. Transforming the first member into a real and an imaginary part (applying cosine and sine to the
angle of 32.28°), we equate to the second member. Using the real part, one finds the value of E, or:
E∠θ = 116.26∠10.45°
Using the imaginary part and applying the function sine to the angle of -32.28°, we find a
negative value, meaning that the X reactance must be a capacitive reactance. Then:
((116.26 - 68) / 20.4) sen(-32.28°) = X
Carrying out the calculation:
X = - 1.26 Ω
Therefore, the X element is a capacitor that must have a reactance of 1.26 Ω.
