Question 5.4: Steady-State AC Analysis of a Series Circuit Find the steady...

From the expression given for the source voltage v_s(t), we see that the peak voltage is 100 V, the angular frequency is ω = 500, and the phase angle is 30°.
The phasor for the voltage source is

 \mathrm{V_s}=100\angle 30^\circ

The complex impedances of the inductance and capacitance are

\mathrm{Z}_L =j\omega L=j500\times 0.3=j150~\Omega

and

\mathrm{Z}_C=-j\frac{1}{\omega C}=-j\frac{1}{500\times 40\times 10^{-6}}=-j50~\Omega

The transformed circuit is shown in Figure 5.12(b). All three elements are in series. Thus, we find the equivalent impedance of the circuit by adding the impedances of all three elements:

\mathrm{Z_{eq}}=R+\mathrm{Z}_L+\mathrm{Z}_C

Substituting values, we have

\mathrm{Z_{eq}}= 100+j150-j50=100+j100

Converting to polar form, we obtain

\mathrm{Z_{eq}}= 141.4\angle 45^\circ

Now, we can find the phasor current by dividing the phasor voltage by the equivalent impedance, resulting in

\mathrm{ I }=\frac{\mathrm{V}_s}{Z}=\frac{100\angle 30^\circ}{141.4\angle 45^\circ}=0.707\angle -15^\circ

As a function of time, the current is

i(t)=0.707\cos{(500t-15^\circ)}

Next, we can find the phasor voltage across each element by multiplying the phasor current by the respective impedance:

\mathrm{V}_R=R\times \mathrm{I}=100\times 0.707\angle -15^\circ =70.7\angle -15^\circ

\mathrm{V}_L=j\omega L\times \mathrm{I}=\omega L\angle 90^\circ \times \mathrm{I}=150\angle 90^\circ \times 0.707 \angle -15^\circ

\mathrm{V}_C=-j\frac{1}{\omega C}\times \mathrm{I}=\frac{1}{\omega C}\angle -90^\circ \times \mathrm{I}=50\angle -90^\circ \times 0.707\angle -15^\circ =35.4\angle -105^\circ

The phasor diagram for the current and voltages is shown in Figure 5.13. Notice that the current I lags the source voltage V_s by 45°. As expected, the voltage V_R and current I are in phase for the resistance. For the inductance, the voltage V_L leads the current I by 90°. For the capacitance, the voltage V_C lags the current by 90°.