Calculate the following: The current flowing through each branch, The total current, The total impedance, The phase angle of the circuit, Draw the phasor diagram.

Haya, twende na swali la 2.1.

Data uliyopewa:

• Resistor, $R = 50 \Omega$
• Inductor, $L = 0.3$ H
• Capacitor, $C = 60 \mu\text{F} = 60 \times 10^{-6}$ F
• Voltage ya supply, $V = 200$ V (RMS)
• Frequency ya supply, $f = 50$ Hz

Hatua ya 1: Kokotoa frequency ya angular na reactances.
Frequency ya angular:

$$\omega = 2\pi f = 2\pi (50 \text{ Hz}) = 100\pi \text{ rad/s} \approx 314.159 \text{ rad/s}$$

Inductive reactance:

$$X_L = \omega L = (100\pi \text{ rad/s})(0.3 \text{ H}) = 30\pi \Omega \approx 94.248 \Omega$$

Capacitive reactance:

$$X_C = \frac{1}{\omega C} = \frac{1}{(100\pi \text{ rad/s})(60 \times 10^{-6} \text{ F})} = \frac{1}{0.006\pi} \Omega = \frac{500}{3\pi} \Omega \approx 53.052 \Omega$$

2.1.1 The current flowing through each branch

• Current kupitia resistor ($I_R$): $$I_R = \frac{V}{R} = \frac{200 \text{ V}}{50 \Omega} = \boxed{4.00 \text{ A}}$$
• Current kupitia inductor ($I_L$): $$I_L = \frac{V}{X_L} = \frac{200 \text{ V}}{30\pi \Omega} = \frac{20}{3\pi} \text{ A} \approx \boxed{2.12 \text{ A}}$$
• Current kupitia capacitor ($I_C$): $$I_C = \frac{V}{X_C} = \frac{200 \text{ V}}{\frac{500}{3\pi} \Omega} = \frac{6\pi}{5} \text{ A} \approx \boxed{3.77 \text{ A}}$$

2.1.2 The total current
Katika mzunguko wa parallel RLC, jumla ya current inapatikana kwa kutumia formula:

$$I_{total} = \sqrt{I_R^2 + (I_C - I_L)^2}$$ $$I_{total} = \sqrt{(4.00 \text{ A})^2 + (3.77 \text{ A} - 2.12 \text{ A})^2}$$ $$I_{total} = \sqrt{16 + (1.65)^2}$$ $$I_{total} = \sqrt{16 + 2.7225}$$ $$I_{total} = \sqrt{18.7225} \approx \boxed{4.33 \text{ A}}$$

2.1.3 The total impedance
Jumla ya impedance ($Z_{total}$) inaweza kukokotolewa kwa kutumia sheria ya Ohm:

$$Z_{total} = \frac{V}{I_{total}} = \frac{200 \text{ V}}{4.33 \text{ A}} \approx \boxed{46.19 \Omega}$$

2.1.4 The phase angle of the circuit
Phase angle ($\phi$) inapatikana kwa kutumia formula:

$$\phi = \arctan\left(\frac{I_C - I_L}{I_R}\right)$$ $$\phi = \arctan\left(\frac{3.77 \text{ A} - 2.12 \text{ A}}{4.00 \text{ A}}\right)$$ $$\phi = \arctan\left(\frac{1.65}{4.00}\right)$$ $$\phi = \arctan(0.4125) \approx \boxed{22.42^\circ}$$

Kwa kuwa $I_C > I_L$, mzunguko ni wa capacitive, na jumla ya current inaongoza voltage.

2.1.5 Draw the phasor diagram.

• Chora voltage $V$ kama reference kwenye positive x-axis.
• Chora current ya resistor $I_R$ ikiwa in-phase na $V$ (kwenye positive x-axis).
• Chora current ya inductor $I_L$ ikiwa inalaga $V$ kwa $90^\circ$ (kwenye negative y-axis).
• Chora current ya capacitor $I_C$ ikiwa inaongoza $V$ kwa $90^\circ$ (kwenye positive y-axis).
• Chora net reactive current $I_C - I_L$ (kwenye positive y-axis, kwa sababu $I_C > I_L$).
• Chora jumla ya current $I_{total}$ kama vector sum ya $I_R$ na $(I_C - I_L)$. Itakuwa katika quadrant ya kwanza, ikiongoza voltage kwa angle ya $22.42^\circ$.

(Kumbuka: Siwezi kuchora mchoro hapa, lakini maelezo hapo juu yanaeleza jinsi ya kuuchora.)