In the series-parallel RC circuit of Figure 15-51, determine the following:
(a) total impedance (b) total current (e) phase angle by which $I_{tot}$ leads $V_{s}$

Question

In the series-parallel RC circuit of Figure 15-51, determine the following:
(a) total impedance (b) total current (e) phase angle by which [latex]I_{tot}[/latex] leads [latex]V_{s}[/latex]

Accepted Answer

(a) First, calculate the magnitudes of capacitive reactance.

[latex]X_{C1}= \frac{1}{2\pi fC} = \frac{1}{2\pi (5 \ kHz)(0.1 \ \mu F)}= 318 \ \Omega[/latex]

[latex]X_{C2}= \frac{1}{2\pi fC} = \frac{1}{2\pi (5 \ kHz)(0.047 \ \mu F)}= 677 \ \Omega[/latex]

One approach is to find the impedance of the series portion and the impedance of the parallel portion and combine them to get the total impedance. The impedance of the series combination of [latex]R_{1}[/latex] and [latex]C_{1}[/latex] is

[latex]Z_{1}[/latex] =[latex]R_{1}[/latex] - j[latex]X_{C1}[/latex] = 1.0 kΩ - j318 Ω

To determine the impedance of the parallel portion, first determine the admittance of the parallel combination of [latex]R_{2}[/latex] and [latex]C_{2}[/latex]

[latex] G_{2}= \frac{1}{R_{2}}= \frac{1}{680 \ \Omega}= 1.47 \ mS[/latex]

[latex] B_{C2}= \frac{1}{X_{C2}}= \frac{1}{677 \ \Omega}= 1.48 \ mS[/latex]

[latex]Y_{2}[/latex] = [latex]G_{2}[/latex] + j [latex] B_{C2}[/latex] = 1.47 mS + j1.48 mS

Converting to polar form yields

[latex] Y_{2}= \sqrt{G^{2}_{2} + B^{2}_{C2}} \angle an ^{-1}\left(\frac{B_{C2}}{G_{2}} ight)[/latex]

[latex]\ \ \ \ = \sqrt{(1.47 \ mS)^{2} + (1.48 \ mS)^{2}} \angle an ^{-1}\left(\frac{1.48 \ mS}{1.47 \ mS} ight)= 2.09 \angle 45.2°mS[/latex]

Then, the impedance of the parallel portion is

[latex]Z_{2}= \frac{1}{Y_{2}}=\frac{1}{(2.09 \angle 45.2° mS)}= 478 \angle -45.2° \ \Omega[/latex]

Converting to rectangular form yields

[latex]Z_{2}= Z_{2}\cos heta - jZ_{2}\sin heta [/latex]

[latex]\ \ \ \ \ \ = (478 \ \Omega) \cos (-45.2°)- j(478 \ \Omega) \sin (-45.2°)= 337 \ \Omega - j339 \ \Omega[/latex]

The series portion and the parallel portion are in series with each other. Combine [latex]Z_{1}[/latex] and [latex] Z_{2}[/latex] to get the total impedance.

[latex]Z_{tot} = Z_{1} + Z_{2}[/latex]

[latex]\ \ \ \ \ \ = (1.0 \ k\Omega -j318 \ \Omega) + (337 \ \Omega - j339 \ \Omega) = 1337 \ \Omega - j657 \ \Omega[/latex]

Expressing [latex]Z_{tot}[/latex] in polar form yields

[latex]Z_{tot} =\sqrt{Z^{2}_{1}+ Z^{2}_{2}} \angle - an ^{-1}\left(\frac{Z_{2}}{Z_{1}} ight)[/latex]

[latex]\ \ \ \ \ \ = \sqrt{(1338 \ \Omega )^{2}+ (657 \ \Omega )^{2}} \angle - an ^{-1}\left(\frac{657 \ \Omega}{1337 \ \Omega} ight)= 1.49 \angle -26.2° k\Omega [/latex]

(b) Use Ohm's law to determine the total current.

[latex]I_{tot}= \frac{V_{s}}{Z_{tot}}= \frac{10 \angle 0° V}{1.49 \angle -26.2° k\Omega} = 6.71 \angle 26.2° mA[/latex]

(c) The total current leads the applied voltage by 26.2°.

Question 15.21: In the series-parallel RC circuit of Figure 15-51, determine...

Related Answered Questions