Problems Resolved on R-C Circuits in AC

Problem 53-3 Source: Problem elaborated by the author of the site.

In the Figure 53-03.1, where V = 100 ∠ 0°, determine:

a) the current I.

b) the voltage, V_b, at the point b.

c) the voltage, V_c, at the point c.

Solution of the Problem 53-3

Item a

Note that the capacitor connected to the point c is in series with the resistor of 20 ohms that interconnects the b-c points. Therefore, they form a impedance that will be called of Z_s, equal to Z_s = 20 - j5 = 21.61 ∠-14.04°. On the other hand, this impedance is in parallel with the capacitor of value equal to - j20. Then, we can calculate the impedance at point b, which will be called of Z_b. Then:

Z_b = Z_s (- j20) / Z_s + (- j20)

Replacing Z_s by its numerical value:

Z_b = (21.61 ∠-14.04° ) 20 ∠-90° / (20 - j5 - j20 )

Note that the - j20 = 20 ∠-90° transformation was used in the above equation. By doing calculation is determined for Z_b the value of:

Z_b = 12.88 ∠-52.7° = 7.81 - j10.25

Between the a-b points there is the parallel of the 10 ohms resistor with the capacitor - j10. This impedance shall be Z_a-b and its value is:

Z_a-b = 10 x 10 ∠-90° / 10 - j10 = 7.07 ∠-45° = 5 - j5

By the circuit, it is noticed that Z_a-b and Z_b are in series. Calculating this series determines the value of the total impedance that the circuit offers to the voltage source. Soon:

Z_total = Z_a-b + Z_b = 7.81 - j10.25 + 5 - j5

By performing the calculation, we will find Z_total the value of:

Z_total = 12.81 - j15.25 = 19.92 ∠-50°

Knowing the value of Z_total can calculate the value of I:

I = V / Z_total = 100 ∠0° / 19.92 ∠-50° = 5.02 ∠+50°

Notice that this angle of + 50°, shows that the electric current in the circuit is advanced by 50° with respect to the electrical voltage of the source feeds the circuit. This characterizes a circuit with capacitive predominance.

Item b

To calculate the voltage at the point b simply make the product between the impedance at that point and the current I. Like this:

V_b = Z_b I = 12.88 ∠-52.7° x 5.02∠50° = 64.66 ∠-2.7° V

Item c

In order to calculate the voltage at the point c it is necessary to know the current that circulates by the series circuit that we call the item a of Z_s = 21.61 ∠-14.04°. As we know V_b, applying Ohm's law results:

I_s = V_b / Z_s = 64.66 ∠-2.7° / 21.61 ∠-14.04°

By performing the calculation, we find to I_s the value of:

I_s = 3 ∠+11.34° A

Therefore, V_c will be the product between I_s and the capacitor connected to the point c, or:

V_c = I_s (-j5) = 3 ∠+11.34° x 5 ∠-90° = 15 ∠-78.66° V

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