Problems Resolved on RLC Circuits in AC

Problem 55-1 Source: Problem 9.27 - page 263 - NILSSON & RIEDEL - Book: Circuitos Elétricos - 8ª edição - Ed. Pearson - 2010.

In the Figure 55-1.1, it is known that, R₁ = 240 ohms , R₂ = 80 ohms , C = 2.5 uF , L = 48 mH. The value of the source is I_N = 200 cos (5000t) mA.

a) Determine the total impedance of the circuit and V_N.

b) Determine i_C , i_L and V_C

c) Make a phasor diagram of the circuit.

Solution of the Problem 54-1

Item a

From the equation of I_N we determine the value of ω:

ω = 5 000 rad/s

We can calculate the value of the inductive and capacitive reactance, that is:

X_L = ω L = 5 000 x 48 x 10^-3 = 240 ohms

X_C = 1 / ω C = 1 / (5 000 x 2.5 x 10^-6 ) = 80 ohms

Let two impedances be defined by:

Z_L = R₂ + j X_L = 80 + j 240 = 253 ∠+71.57°

Z_C = R₁ - j X_C = 240 - j 80 = 253 ∠-18.43°

To obtain the total impedance of the circuit, simply calculate the parallel of Z_L and Z_C, or:

Z_T = Z_L Z_C / Z_L + Z_C = 64 009∠53.14° / 357.77∠26.57°

Then, by performing the calculation:

Z_T = 160 + j 80 = 179∠26.57°

What does this value mean Z_T ? Well, that means we could replace the impedances Z_L and Z_C by a single circuit consisting of a resistor with a value equal to 160 ohms in series with an inductor, where L = 16 mH, which has an inductive reactance of 80 ohms when ω = 5 000 rad/s.

Knowing Z_T and I_N we can calculate the voltage V_N on the current source.

V_N = I_N Z_T = 200 ∠0° x 179∠26.57°

V_N = 35 800∠26.57° mV = 35.8∠26.57° V

Keep in mind that this value of V_N is the peak value of the voltage on the current source I_N. Thus, in the calculation of i_L and i_C, then the values found will also be peak values, since we use the peak value in V_N for the calculation of these currents.

Item b

With this value of V_N, we easily calculate the values of i_L and i_C. Then:

i_L = V_N / Z_L = 35.8∠26.57°/ 253∠71.57° = 0.142∠-45° A

i_C = V_N / Z_C = 35.8∠26.57°/ 253∠-18.43° = 0.142∠+45° A

It is now easy to calculate the V_C.

V_C = X_C i_C = 80∠-90° x 0.142∠45° = 11.36∠-45° V

With this value of i_C, we can also calculate the value of V_R1, or:

V_R1 = R₁ i_C = = 240 x 0.142∠45° = 34.08∠+45° V

Note that the voltage phase on the capacitor (V_C ) is of - 45° while that of the current (i_C ) is of +45° confirming that in a capacitor the current is advanced by 90° in relation to its voltage. In the inductor the same happens, but the current is delayed by 90° in relation to its voltage. Check !!!!!

Item c

In the Figure 55-1.2 we show the complete phasor diagram of the circuit. It is evident that I_N is the phasor sum of i_L with i_C. Similarly, V_N is the phasor sum, both in V_R1 with V_C, as also V_R2 with V_L, since the circuits are in parallel. Because of this, we can apply the Pythagorean theorem directly, obtaining:

|I_N|² = |i_L|² + |i_C|²

|V_N|² = |V_R1|² + |V_C|²

|V_N|² = |V_R2|² + |V_L|²

Back to Top

Previous Problem

Next Problem