Find the current i(t) in the network in Fig. E8.8.Figure E8.8.

The impedance of the inductance is as follows:

$$ Z_{L}=j \omega L $$

Calculate the inductance impedance by substituting \(60 \mathrm{~Hz}\) for \(f\) and \(40 \mathrm{mH}\) for \(L\).

$$ \begin{aligned} Z_{L} &=j \omega L \\ &=j(2 \pi)(60)\left(40 \times 10^{-3}\right) \\ &=j 15.07 \Omega \end{aligned} $$

The impedance of the capacitance is as follows:

$$ Z_{C}=\frac{-j}{\omega C} $$

Calculate the capacitance impedance by substituting \(60 \mathrm{~Hz}\) for \(f\) and \(50 \mu \mathrm{H}\) for \(C\).

$$ \begin{aligned} Z_{C} &=\frac{-j}{\omega C} \\ &=\frac{-j}{2 \pi(60)\left(50 \times 10^{-6}\right)} \\ &=-j 53.07 \Omega \end{aligned} $$

Write the expression for the equivalent impedance.

$$ Z=\left(Z_{R}+Z_{L}\right) \| Z_{C} $$

Calculate the equivalent impedance.

Substitute \(20 \Omega\) for \(Z_{R}, j 15.07 \Omega\) for \(Z_{L}\) and \(-j 53.07 \Omega\) for \(Z_{C}\) in the equation.

$$ \begin{aligned} Z &=\left(Z_{R}+Z_{L}\right) \| Z_{C} \\ &=(20+j 15.07) \|(-j 53.07) \\ &=\frac{799.76-j 1061.4}{20-j 38} \\ &=\frac{1328.97 \angle-53^{\circ}}{42.94 \angle-62.24^{\circ}} \\ &=30.94 \angle 9.24^{\circ} \Omega \end{aligned} $$

The voltage source is as follows:

$$ \begin{aligned} v(t) &=120 \sin \left(377 t+60^{\circ}\right) \mathrm{V} \\ &=120 \cos \left(377 t+60^{\circ}-90^{\circ}\right) \mathrm{V} \\ &=120 \cos \left(377 t-30^{\circ}\right) \mathrm{V} \\ &=120 \angle-30^{\circ} \mathrm{V} \end{aligned} $$

Write the expression for the current in the circuit.

$$ \mathrm{I}=\frac{\mathrm{V}}{\mathrm{Z}} $$

Calculate the current in the circuit.

Substitute \(120 \angle-30^{\circ}\) for \(\mathrm{V}\) and \(30.94 \angle 9.24^{\circ}\) for \(\mathrm{Z}\) in the equation.

$$ \begin{aligned} I &=\frac{V}{Z} \\ &=\frac{120 \angle-30^{\circ}}{30.94 \angle 9.2^{\circ}} \\ &=3.88 \angle-39.2 \mathrm{~A} \end{aligned} $$

Convert the current in phasor form to time domain.

\(i(t)=3.88 \cos \left(377 t-39.2^{\circ}\right) \mathrm{A}\)

Therefore, the current in the circuit is \(3.88 \cos \left(377 t-39.2^{\circ}\right) \mathrm{A}\).