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Chapter 12

Q. 12.15

Repeat Example12.14  for Z=(400-j 300) \Omega .

Step-by-Step

Verified Solution

The impedance  Z  is in the fourth quadrant, so   \tilde{V} leads \tilde{I}  by the angle

∡ Z=\tan ^{-1}\left(-\frac{300}{400}\right)=-0.644<0

or \tilde{V} lags \tilde{I} by  0.644 . For f_0=1 kHz,

\cos \left(\omega_0 t-0.644\right)=\cos \left[\omega_0\left(t-\frac{0.644}{\omega_0}\right)\right]

so \nu(t) lags i(t) by

\frac{0.644}{2 \pi \times 10^3 Hz } \cong 103  \mu s .