Question

A series RLC circuit with R = 15.0 ohms, C = 4.70 uF, and L = 25.0mH is connected to an AC voltage source with V(t) = 75.0 Vrms and frequency = 550 Hz as shown in the figure below.
a) Calculate the rms current in the circuit.
b) Calculate the rms voltages Vab, Vbc, Vcd, Vbd, and Vad.
c) Calculate the average rate at which energy is dissipated in each of the 3 circuit elements.

Answer 1

(a) The rms current in the circuit is 2.58 A.

(b) The rms voltage of Vab is 38.7 V, Vbc is 158.83 V, Vcd is 222.93 V, Vbd is 64.11 V, and Vad is 75 V.

(c) The average rate at which energy is dissipated in each of the 3 circuit elements is 193.23 W.

Capacitive reactance of the circuit

The capacitive reactance of the circuit is calculated as follows;

$X_c = \frac{1}{\omega C} = \frac{1}{2\pi fC} = \frac{1}{2\pi \times 550 \times 4.7 \times 10^{-6}} \\\\X_c = 61.56 \ ohms$

Inductive reactance of the circuit

Xl = ωL

Xl = 2πfL

Xl = 2π x 550 x 25 x 10⁻³

Xl = 86.41 ohms

Impedance of the circuit

$Z = \sqrt{R^2 + (X_L - X_C)^2} \\\\Z = \sqrt{15^2 + (86.41 -61.56)^2 } \\\\Z = 29.03 \ ohms$

rms current in the circuits

$I_{rms} = \frac{V_{rms}}{Z} \\\\I_{rms} = \frac{75}{29.03} \\\\I_{rms} = 2.58 \ A$

rms voltage in resistor (Vab)

$V_{ab} = I_{rms} R\\\\V_{ab} = 2.58 \times 15\\\\V_{ab} = 38.7 \ V$

rms voltage in capacitor (Vbc)

$V_{bc} = I_{rms} X_c\\\\V_{bc} = 2.58 \times 61.56\\\\V_{bc} = 158.83 \ V$

rms voltage in inductor (Vcd)

$V_{cd} = I_{rms} X_l\\\\V_{cd} = 2.58 \times 86.41\\\\V_{cd} = 222.93\ V$

rms voltage in capacitor and inductor (Vbd)

$V_{bd} = I_{rms} \times (X_l - X_c)\\\\V_{bd} = 2.58 \times (86.41 - 61.56)\\\\V_{bd} = 64.11 \ V$

rms voltage in resistor, capacitor and inductor (Vad)

$V_{ad} = I_{rms} \times \sqrt{R^2 + (X_l- X_c)^2} \\\\V_{ad} = 2.58 \times Z \\\\V_{ad} = 2.58 \times 29.03\\\\V_{ad} = 75 \ V$

Average rate of energy dissipation in the 3 circuit element

$P = I_{rms}^2 Z\\\\P = (2.58)^2 \times 29.03\\\\P = 193.23 \ W$

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