Question 15.22: Determine all currents in Figure 15-52. Draw a current phaso...

Determine all currents in Figure 15-52. Draw a current phasor diagram.

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First, calculate X_{C1} and X_{C2}.

X_{C1}= \frac{1}{2\pi fC} = \frac{1}{2\pi (2 \ MHz)(0.001 \ \mu F)}= 79.6 \ \Omega

 

X_{C2}= \frac{1}{2\pi fC} = \frac{1}{2\pi (2 \ MHz)(0.0022 \ \mu F)}= 36.2 \ \Omega

Next, determine the impedance of each of the two parallel branches.

Z_{1} = R_{1} – jX_{C1} = 33 Ω – j79.6 Ω
Z_{2} = R_{2} – j X_{C2} = 47 Ω  – j36.2 Ω

Convert these impedances to polar form.

Z_{1} =\sqrt{R^{2}_{1}+ X^{2}_{C1}} \angle -\tan ^{-1}\left(\frac{X_{C1}}{R_{1}} \right)

 

\ \ \ \ \ = \sqrt{(33 \ \Omega )^{2}+ (79.6 \ \Omega )^{2}} \angle -\tan^{-1}\left(\frac{79.6 \ \Omega}{33 \ \Omega} \right)= 86.2 \angle -67.5° k\Omega

 

Z_{2} =\sqrt{R^{2}_{2}+ X^{2}_{C2}} \angle -\tan ^{-1}\left(\frac{X_{C2}}{R_{2}} \right)

 

\ \ \ \ \ = \sqrt{(47 \ \Omega )^{2}+ (36.2 \ \Omega )^{2}} \angle -\tan^{-1}\left(\frac{36.2 \ \Omega}{47 \ \Omega} \right)= 59.3 \angle -37.6° k\Omega

Calculate each branch current.

I_{1}= \frac{V_{s}}{Z_{1}}= \frac{2 \angle 0° V}{86.2 \angle -67.5° \Omega} = 23.2 \angle 67.5° mA

 

I_{2}= \frac{V_{s}}{Z_{2}}= \frac{2 \angle 0° V}{59.3 \angle -37.6° \Omega} = 33.7 \angle 37.6° mA

To get the total current, express each branch current in rectangular form so that they can be added.

I_{1} = 8.89 mA + j 21.4 mA
I_{2} = 26.7 mA + j 20.6 mA

The total current is

I_{tot} =  I_{1} + I_{2}

 

\ \ \ \ \ \ = (8.89 \ mA + j21.4 \ mA) + (26.7 \ mA + j20.6 \ mA) = 35.6 \ mA + j42.0 \ mA

Converting I_{tot} to polar form yields

I_{tot}= \sqrt{(35.6 \ mA)^{2} + (42.0 \ mA)^{2}} \angle \tan^{-1}\left(\frac{42.0 \ \Omega}{35.6 \ \Omega } \right)= 55.1 \angle 49.7° mA

The current phasor diagram is shown in Figure 15-53.

Screenshot (765)

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