Determine the total current and phase angle in Figure 15—42. Draw a phasor diagram showing the relationship of Vs, and Itot.

Question

Determine the total current and phase angle in Figure 15—42. Draw a phasor diagram showing the relationship of [latex]V_{s}[/latex], and [latex]I_{tot}[/latex].

Accepted Answer

The capaeitive reactance is

[latex]X_{C}= \frac{1}{2\pi fC} = \frac{1}{2\pi (1.5 \ kHz) (0.022 \ \mu F)}= 4.82 \ k\Omega[/latex]

The capaeitive susceptance magnitude is

[latex]B_{C}= \frac{1}{X_{C}}= \frac{1}{4.82 \ k\Omega }= 207 \ \mu S[/latex]

The conductance magnitude is

[latex]G= \frac{1}{R}= \frac{1}{2.2 \ k\Omega }= 455 \ \mu S[/latex]

The total admittance is

[latex]Y_{tot}[/latex]= G + j[latex]B_{C}[/latex] = 455 [latex]\mu[/latex] S + j 207 [latex]\mu[/latex] S

Converting to polar form yields

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

[latex]\ \ \ \ \ \ =\sqrt{(455 \ \mu S)^{2}+ (207 \ \mu S)^{2} }\angle an ^{-1}\left(\frac{207 \ \mu S}{455 \ \mu S} ight) = 500 \angle 24.5° \mu S[/latex]

The phase angle is 24.5°.
Use Ohm's law to determine the total current.

[latex]I_{tot}= V_{s}Y_{tot}= (10\angle 10° V)(500\angle 24.5° \mu S)= 5.00 \angle 25.5° mA[/latex]

The magnitude of the total current is 5.00 mA, and it leads the applied voltage by 24.5°, as the phasor diagram in Figure 15-43 indicates.

Question 15.18: Determine the total current and phase angle in Figure 15—42....

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