Problems Resolved on RLC Circuits in AC

Problem 55-2 Source: Problem 5 - page 510 - BOYLESTAD, Robert - Book: Introdução à Análise de Circuitos - Ed. Pearson Education do Brasil - 10ª edição - 2004.

For the circuit shown in the Figure 55-2.1, calculate:

a) the current I.

b) the voltage V_C.

c) the average power supplied to the circuit.

Solution of the Problem 54-2

Item a

Note that the two capacitors of -j 400 each, are in parallel. Thus, the equivalent capacitance is equal to -j 200. The impedance composed of the R₂ and the capacitors will be called of Z_C, which value is:

Z_C = 200 - j200 = 200 √2 ∠-45° Ω

On the other hand, the same can be done for the other impedance, calling it Z_L, or:

Z_L = 560 + j560 = 560 √2 ∠+45° Ω

We can calculate the values of i_C and i_L.

i_C = V / Z_C = 100 / 200 √2 ∠-45°

Performing the calculation:

i_C = 0.35 ∠+45° A

To i_L:

i_L = V / Z_L = 100 / 560 √2 ∠+45°

Performing the calculation:

i_L = 0.13 ∠-45° A

Directly from the circuit, it is known that I = i_C + i_L. Then:

I = i_C + i_L = 0.35 ∠+45° + 0.13 ∠-45°

Performing the calculation:

I = 0.339 + j0.155 = 0.37 ∠+24.62° A

Notice that the current I is advanced of 24.62° with respect to the voltage, meaning that we have a circuit with capacitive predominance.

Item b

To find the value of V_C, the Ohm's law is used. Like this:

V_C = - j200 i_C = 200 ∠-90° x 0.35 ∠+45°

Performing the calculation:

V_C = 70 ∠-45° V

Item c

In order to calculate the average power supplied to the circuit by the source, we use the equation:

P = |V| |I| cos φ = 100 x 0.37 x cos 24.62°

Performing the calculation:

P = 33.64 watts

Back to Top

Previous Problem

Next Problem