Tentukan nilai impedansi keseluruhan (overall impedance) untuk kasus-kasus berikut:
(a) $P = 1000 \text{ W}$, $pf = 0.8 \ (\text{leading})$, $V_{rms} = 220 \text{ V}$
(b) $P = 1500 \text{ W}$, $Q = 2000 \text{ VAR} \ (\text{inductive})$, $I_{rms} = 12 \text{ A}$
(c) $\mathbf{S} = 4500\angle60^\circ \text{ VA}$, $\mathbf{V} = 120\angle45^\circ \text{ V}$
Ada tiga rumus utama yang menghubungkan Daya Kompleks ($\mathbf{S}$) dengan Impedansi ($\mathbf{Z}$):
Jika diketahui $V_{rms}$:
$$\mathbf{Z}^* = \frac{|\mathbf{V_{rms}}|^2}{\mathbf{S}} \quad \text{atau} \quad \mathbf{Z} = \frac{|\mathbf{V_{rms}}|^2}{\mathbf{S}^*}$$
Jika diketahui $I_{rms}$:
$$\mathbf{Z} = \frac{\mathbf{S}}{|\mathbf{I_{rms}}|^2}$$
Sifat Sudut:
- Leading (Kapasitif): Sudut impedansi ($\theta_z$) harus negatif.
- Lagging/Inductive: Sudut impedansi ($\theta_z$) harus positif.
(a) $P = 1000 \text{ W}$, $pf = 0.8 \ (\text{leading})$, $V_{rms} = 220 \text{ V}$
Kita harus mencari daya semu
$$S = \frac{P}{pf}$$
$$S = \frac{1000}{0.8}$$
$$S = 1250 \text{ VA}$$
Kita harus mencari sudut daya
$$\theta = \cos^{-1}(pf)$$
$$\theta = \cos^{-1}(0.8)$$
$$\theta = 36.869^\circ$$
Kita bisa menyusun daya kompleksnya
$$S = 1250 \angle 36.869^\circ$$
Kita ubah menjadi konjugasi
$$S = 1250 \angle 36.869^\circ$$
Kita tinggal mencari impedansinya
$$Z = \frac{|V_{rms}|^2}{S^*}$$
$$Z = \frac{220^2}{1250 \angle 36.869^\circ}$$
$$Z = \frac{48400}{1250 \angle 36.869^\circ}$$
$$Z = \frac{48400}{1250} \angle (0^\circ – 36.869^\circ)$$
$$\mathbf{Z = 38.72 \angle -36.869^\circ \ \Omega}$$
(b) $P = 1500 \text{ W}$, $Q = 2000 \text{ VAR} \ (\text{inductive})$, $I_{rms} = 12 \text{ A}$
Kita bisa menyusun daya kompleksnya
$$S = P + jQ$$
$$S = 1500 + j2000 \text{ VA}$$
Kita ubah daya kompleks menjadi bentuk fasor
$$\sqrt{1500^2 + 2000^2} = \sqrt{2250000 + 4000000} = \sqrt{6250000} = 2500$$
$$\tan^{-1} = (\frac{2000}{1500}) = \tan^{-1}(1.333) = 53.13$$
$$S = 2500 \angle 53.13^\circ$$
Kita ubah menjadi konjugasi
$$S = 2500 \angle -53.13^\circ$$
Kita tinggal mencari impedansinya
$$Z = \frac{S^*}{|I_{rms}|^2}$$
$$Z = \frac{2500 \angle -53.13^\circ}{12^2}$$
$$Z = \frac{2500 \angle -53.13^\circ}{144}$$
$$Z = \frac{2500}{144} \angle (0^\circ + 53.13^\circ)$$
$$\mathbf{Z = 17.361 \angle 53.13^\circ \ \Omega}$$
(c) $\mathbf{S} = 4500\angle60^\circ \text{ VA}$, $\mathbf{V} = 120\angle45^\circ \text{ V}$
Kita ubah menjadi konjugasi
$$S = 4500 \angle -60^\circ$$
Kita tinggal mencari impedansinya
$$Z = \frac{|V_{rms}|^2}{S^*}$$
$$Z = \frac{120^2}{4500 \angle 60^\circ}$$
$$Z = \frac{14400}{4500 \angle 60^\circ}$$
$$Z = \frac{14400}{4500} \angle (0^\circ – 60^\circ)$$
$$\mathbf{Z = 3.2 \angle -60^\circ \ \Omega}$$
