A 90 Ω resistor, a 32 mH inductor, and a 5μF capacitor are connected in series across the terminals of a sinusoidal voltage source, as shown in Fig. 9.15. The steady-state expression for the source voltage v_{s} is 750\cos (5000t + 30^{°} ) V.
a) Construct the frequency-domain equivalent circuit.
b) Calculate the steady-state current i by the phasor method.
a) From the expression for v_{s} , we have \omega=5000\frac{rad}{s}.Therefore the impedance of the 32 mH inductor is
Z_{L} = j\omega L = j\left(5000\right)\left(32 \times 10^{-3}\right) = j160 \Omega,
and the impedance of the capacitor is
Z_{C} = j\frac{-1}{\omega C}= -j\frac{10^{6}}{\left(5000\right)\left(5\right) }= -j40 \Omega.
The phasor transform of v_{s} is
V_{s}= 750 \angle 30^{°} V.
Figure 9.16 illustrates the frequency-domain equivalent circuit of the circuit shown in Fig. 9.15.
b) We compute the phasor current simply by dividing the voltage of the voltage source by the equivalent impedance between the terminals a, b. From Eq. 9.45 Z_{ab }=\frac{V_{ab}}{I}=Z_{1}+Z_{2}+…+Z_{n},
Z_{ab }= 90 + j160 – j40= 90 + j120 = 150\angle 53.13^{° }\Omega.
Thus
I =\frac{750 \angle 30^{°}}{150\angle53.13^{°}}= 5 \angle -23.13^{°}A.
We may now write the steady-state expression for i directly:
i=5\cos (5000t + 23.13^{° }) A.
