Question 5.5: Series and Parallel Combinations of Complex Impedances Consi...

The phasor for the voltage source is \mathrm{V}_s=10\angle -90^\circ. [Notice that v_s(t) is a sine function rather than a cosine function, and it is necessary to subtract 90° from the phase.] The angular frequency of the source is ω = 1000. The impedances of the inductance and capacitance are

Z_L=j\omega L=j1000\times 0.1=j100~\Omega

and

Z_C =-j\frac{1}{\omega C } =-j\frac{1}{1000\times 10 \times10^{-6}}=-j100~\Omega

The transformed network is shown in Figure 5.14(b).

To find V_C, we will first combine the resistance and the impedance of the capacitor in parallel. Then, we will use the voltage-division principle to compute the voltage across the RC combination. The impedance of the parallel RC circuit is

Z_{RC}=\frac{1}{1/R+1/Z_C}=\frac{1}{1/100+1/(-j100)} =\frac{1}{0.01+j0.01}=\frac{1\angle0^\circ }{0.01414\angle 45^\circ }=70.71\angle -45^\circ  

Converting to rectangular form, we have

Z_{RC}=50-j50

The equivalent network is shown in Figure 5.14(c).

Now, we use the voltage-division principle to obtain

\mathrm{V}_C=\mathrm{V}_s\frac{Z_{RC}}{Z_L+Z_{RC}}=10\angle -90^\circ \frac{70.71\angle -45^\circ }{j100+50-j50}=10\angle -90^\circ \frac{70.71\angle -45^\circ}{50+j50}=10\angle -90^\circ\frac{70.71\angle -45^\circ}{70.71\angle 45^\circ} =10\angle -180^\circ

Converting the phasor to a time function, we have

v_C(t)=10\cos{(1000t-180^\circ)}=-10\cos {(1000t)}

Next, we compute the current in each element yielding

\mathrm{I}=\frac{\mathrm{V}_s}{Z_L+Z_{RC}}=\frac{10\angle -90^\circ}{j100+50-j50}=\frac{10\angle -90^\circ}{50+j50}=\frac{10\angle -90^\circ}{70.71\angle 45^\circ} =0.1414\angle-135^\circ

\mathrm{I}_R=\frac{\mathrm{V}_C}{R}=\frac{10\angle -180^\circ}{100}=0.1\angle -180 ^\circ

\mathrm{I}_C=\frac{\mathrm{V}_C}{Z_C}=\frac{10\angle -180^\circ}{-j100}= \frac{10\angle -180^\circ}{100\angle -90^\circ}=0.1\angle -90 ^\circ

The phasor diagram is shown in Figure 5.15.