Problem + Hard 55-4
Source: Problem 38 - 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 we have to |VR| = 2 |I|. We know that VR is
delayed 90° with respect to I. Furthermore, X is a component
R , L or pure C. Determine R and X.
Solution of the Problem + Hard 55-4
Let's try to simplify the circuit solving the parallel of the impedances 2 + j2 and 2 - j2.
Doing the calculation we find the equivalent of a resistor with a value equal to 2 ohms. On the other hand,
on the right side of the circuit, we can add the -j reactance of the capacitor and the 4 ohms resistor with the impedance that is inside the red rectangle, as these components are in series.
In the figure below we can see how the circuit turned out after simplification.
According to the circuit, we can calculate the value of I, or:
I = E∠0° / (2 - j4) = 0.224 E∠63.43°
Notice that 2 - j 2 = 4.472∠-63.43°. But from the statement of the problem we know that
|VR| = 2 |I| and also that VR is 90° behind by
relation to I. Soon:
VR = 2 I = 2 x 0.224 E∠(63.43° - 90°) = 0.448 E∠-26.57°
For the circuit, VR is the voltage across the resistance equal to 4 ohms.
Then we can calculate the current flowing through it, which we will call I1, as
circuit above.
I1 = VR / 4 = 0.112 E∠-26.57°
Knowing the current I1, the equivalent impedance of the circuit can be calculated on the right of the
resistor of 4 ohms (including it), that is, between points a - c, which will be called
Zac. Then:
Zac = E∠0° / I1 = 8.9286∠26.57 = 8 + j 4 Ω
Now, subtracting from this value of Zac the value of the impedance that is inside the red rectangle, we find the
value of the impedance that the circuit formed by paralleling the impedances R ± jX and 2 + j4, or
that is, the impedance between points b - c and which will be called Zbc. Then:
Zbc = 8 + j 4 - (4.8 + j 1.6) = 3.2 + j 2.4
To calculate R and X, calculate the parallel of R ± jX and 2 + j 4 and equals the value of
Zbc found above. Like this:
Algebraically working the above relation, equating the real part of the first member with that of the second member and doing the same with the imaginary part, we arrive at a system of two equations with two unknowns, as follows:
1.2 R + 1.6 X = 3.2
1.6 R - 1.2 X = 17.6
And the solution of this system is:
R = 8 and X = -4
Therefore, it is concluded that the value of R is 8 ohms and X is a capacitor that presents
4 ohms reactance.
