Sebuah hambatan \(3 \text{ Ohm}\) dan kumparan \(L = 40 \text{ mH}\) dihubungkan pada sumber arus bolak balik dengan tegangan efektif \(120 \text{ Volt}\) dan frekuensi \(\dfrac {50}{\pi} \text{ Hz}\).
(A) Impedansi rangkaian
Reaktansi induktif
\begin{equation*}
\begin{split}
X_L & = \omega \:.\: L \\\\
X_L & = 2\pi \:.\: f \:.\: L \\\\
X_L & = 2\pi \:.\: \frac {50}{\pi} \:.\: 40 \:.\: 10^{-3} \\\\
X_L & = 4 \: \Omega
\end{split}
\end{equation*}
Impedansi
\begin{equation*}
\begin{split}
Z & = \sqrt{R^2 + X_L^2} \\\\
Z & = \sqrt{3^2 + 4^2} \\\\
Z & = 5 \: \Omega
\end{split}
\end{equation*}
(B) Kuat arus efektif
\begin{equation*}
\begin{split}
V & = I \:.\: Z \\\\
120 & = I \:.\: 5 \\\\
I & = 24 \text{ A}
\end{split}
\end{equation*}
(C) Beda fase antara tegangan dan arus listrik
\begin{equation*}
\begin{split}
\tan \theta & = \frac {X_L - X_C}{R} \\\\
\tan \theta & = \frac {4 - 0}{3} \\\\
\tan \theta & = \frac {4}{3} \\\\
\theta & = 53^{\text{o}}
\end{split}
\end{equation*}
Tegangan mendahului arus 53º
(D) Persamaan tegangan dan kuat arus
\(V_{\text{max}} = V_{\text{eff}} \:.\: \sqrt{2} = 120 \sqrt{2}\)
\(I_{\text{max}} = I_{\text{eff}} \:.\: \sqrt{2} = 24 \sqrt{2}\)
\(V = 120 \sqrt{2} \:.\: \sin (100 t + 53)^{\text{o}}\)
\(I = 24 \sqrt{2} \:.\: \sin (100 t)^{\text{o}}\)
(E) Tegangan pada ujung-ujung resistor dan induktor
\begin{equation*}
\begin{split}
V_R & = I \:.\: R \\\\
V_R & = 24 \:.\: 3 \\\\
V_R & = 72 \text{ V}
\end{split}
\end{equation*}
\begin{equation*}
\begin{split}
V_L & = I \:.\: X_L \\\\
V_L & = 24 \:.\: 4 \\\\
V_L & = 96 \text{ V}
\end{split}
\end{equation*}