Let’s solve the given problem step-by-step based on the circuit and questions presented in the image.

1. Problem Summary

We are working with a series RLC circuit:

• Resistance, $R = 3 \, \Omega$
• Inductive reactance, $X_L = j6 \, \Omega$
• Capacitive reactance, $X_C = -j2 \, \Omega$
• Voltage source, $V = 60 \angle 0^\circ \, \text{V}$

(a) Calculate total impedance $Z_T$ and current $I_T$:

Total Impedance:

$Z_T = R + j(X_L + X_C)$ Substitute values: $Z_T = 3 + j(6 - 2) = 3 + j4 \, \Omega$

Convert $Z_T$ to polar form: $|Z_T| = \sqrt{R^2 + X^2} = \sqrt{3^2 + 4^2} = 5 \, \Omega$ $\theta_Z = \tan^{-1}\left(\frac{X}{R}\right) = \tan^{-1}\left(\frac{4}{3}\right) = 53.13^\circ$ Thus: $Z_T = 5 \angle 53.13^\circ \, \Omega$

Current $I_T$:

Using Ohm's law: $I_T = \frac{V}{Z_T}$ $I_T = \frac{60 \angle 0^\circ}{5 \angle 53.13^\circ} = 12 \angle -53.13^\circ \, \text{A}$

(b) Phasor diagram of voltage $V$ and current $I_T$:

The voltage $V$ is at $0^\circ$, and the current $I_T$ lags behind $V$ by $53.13^\circ$. The phasor diagram will show:

• $V$ as a horizontal vector along the real axis.
• $I_T$ as a vector at an angle of $-53.13^\circ$ from $V$.

(c) Determine:

(i) Real Power ($P$):

$P = V_\text{rms} I_\text{rms} \cos\theta_Z$ $P = \frac{60}{\sqrt{2}} \cdot \frac{12}{\sqrt{2}} \cdot \cos(53.13^\circ)$ $P = 30 \cdot 6 \cdot 0.6 = 108 \, \text{W}$

(ii) Reactive Power ($Q$):

$Q = V_\text{rms} I_\text{rms} \sin\theta_Z$ $Q = \frac{60}{\sqrt{2}} \cdot \frac{12}{\sqrt{2}} \cdot \sin(53.13^\circ)$ $Q = 30 \cdot 6 \cdot 0.8 = 144 \, \text{VAR}$

(iii) Apparent Power ($S$):

$S = V_\text{rms} I_\text{rms}$ $S = \frac{60}{\sqrt{2}} \cdot \frac{12}{\sqrt{2}} = 30 \cdot 6 = 180 \, \text{VA}$

(iv) Phase Angle ($\phi$):

$\phi = \theta_Z = 53.13^\circ$

(d) Draw the Power Triangle:

The power triangle will:

• Have $P = 108 \, \text{W}$ on the horizontal axis.
• Have $Q = 144 \, \text{VAR}$ on the vertical axis.
• The hypotenuse will represent $S = 180 \, \text{VA}$.
• The angle between $P$ and $S$ is $\phi = 53.13^\circ$.

(e) Power Factor:

The power factor is: $\text{Power Factor} = \cos\phi = \cos(53.13^\circ) = 0.6$ Since the current lags the voltage, it is lagging.

Would you like me to plot the phasor diagram or power triangle? Or explain any of these steps in more detail?

Related Questions:

1. How does the phase angle affect the power factor in RLC circuits?
2. What are the implications of a lagging power factor in power systems?
3. How can the impedance of the circuit be reduced to minimize reactive power?
4. How do capacitors and inductors impact the power triangle?
5. What would happen if $X_L = X_C$ in this circuit?

Tip:

Always convert impedance and current into their polar forms to simplify AC circuit calculations.