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CHAP. 9]

203

SINUSOIDAL STEADY-STATE CIRCUIT ANALYSIS

By the methods of Example 9.1, 500ð20  103 Þ ¼ 458  ¼ arctan 10 qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi v ¼ I R2 þ ð!LÞ2 sin ð!t þ Þ ¼ 28:3 sin ð500t þ 458Þ

ðVÞ

It is seen that i lags v by 458.

9.3

Find the two elements in a series circuit, given that the current and total voltage are i ¼ 10 cos ð5000t  23:138Þ ðAÞ

v ¼ 50 cos ð5000t þ 308Þ

ðVÞ

Since i lags v (by 53.138), the elements are R and L. The ratio of Vmax to Imax is 50/10. qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 50 5000L and tan 53:138 ¼ 1:33 ¼ ¼ R2 þ ð5000LÞ2 10 R

Hence,

Solving, R ¼ 3:0 , L ¼ 0:8 mH.

9.4

A series circuit, with R ¼ 2:0  and C ¼ 200 pF, has a sinusoidal applied voltage with a frequency of 99.47MHz. If the maximum voltage across the capacitance is 24 V, what is the maximum voltage across the series combination? ! ¼ 2f ¼ 6:25  108 rad=s From Table 9-1, Imax ¼ !CVC;max ¼ 3:0 A. Then, by the methods of Example 9.2, qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Vmax ¼ Imax R2 þ ð1=!CÞ2 ¼ ð6Þ2 þ ð24Þ2 ¼ 24:74 V

9.5

The current in a series circuit of R ¼ 5  and L ¼ 30 mH lags the applied voltage by 808. Determine the source frequency and the impedance Z. From the impedance diagram, Fig. 9-16, 5 þ jXL ¼ Z

808

XL ¼ 5 tan 808 ¼ 28:4 

3

Then 28:4 ¼ !ð30  10 Þ, whence ! ¼ 945:2 rad/s and f ¼ 150:4 Hz. Z ¼ 5 þ j28:4 

Fig. 9-16

Fig. 9-17

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