QUESTION 3: AC THEORY 3.1 Name FIVE detrimental effects of a poor power factor. 3.2 The following information about a circuit is known: - \( Z_{1}=5+j 9 \) \( Z_{2}=-8-j 12 \) Determine the following: 3.2.1 \( \quad Z_{1}+Z_{2} \) and convert your answer into the polar form. 3.2.2 \( Z_{1}-Z_{2} \) and convert your answer into the polar form. 3. 3

### **QUESTION 3: AC THEORY** --- #### **3.1 Name FIVE detrimental effects of a poor power factor.** A poor power factor can lead to several adverse effects in an electrical system. Here are five key detrimental impacts: 1. **Increased Power Losses:** - **Explanation:** A low power factor results in higher current flow for the same amount of useful power. This increased current leads to greater I²R (current squared × resistance) losses in electrical conductors and equipment. 2. **Higher Electricity Bills:** - **Explanation:** Many utility companies impose penalties or higher charges for industrial and commercial customers with low power factors. This is because maintaining a low power factor requires utilities to generate and transmit more power, increasing their operational costs. 3. **Reduced System Capacity:** - **Explanation:** With a poor power factor, more current is required to deliver the same amount of real power. This reduces the overall capacity of the electrical system to handle additional loads, limiting the ability to expand or add new equipment without significant upgrades. 4. **Overloading of Electrical Components:** - **Explanation:** Transformers, generators, and other electrical equipment may experience overloading due to the increased current demand. Over time, this can lead to overheating, reduced efficiency, and shortened equipment lifespan. 5. **Voltage Drops and Poor Voltage Regulation:** - **Explanation:** Higher currents associated with a low power factor can cause significant voltage drops across the distribution network. This can result in inadequate voltage levels for sensitive equipment, leading to malfunction or damage. --- #### **3.2 The following information about a circuit is known:** - \( Z_{1} = 5 + j9 \) Ω - \( Z_{2} = -8 - j12 \) Ω --- ##### **3.2.1 \( \mathbf{Z_{1} + Z_{2}} \) and convert your answer into the polar form.** **Step 1: Calculate \( Z_{1} + Z_{2} \)** \[ Z_{1} + Z_{2} = (5 + j9) + (-8 - j12) = (5 - 8) + j(9 - 12) = -3 - j3 \, \Omega \] **Step 2: Convert to Polar Form** The polar form of a complex number \( a + jb \) is given by: \[ |Z| = \sqrt{a^2 + b^2} \] \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \] - **Magnitude (\( |Z| \)):** \[ |Z| = \sqrt{(-3)^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} \approx 4.24 \, \Omega \] - **Phase Angle (\( \theta \)):** \[ \theta = \tan^{-1}\left(\frac{-3}{-3}\right) = \tan^{-1}(1) = 45^\circ \] Since both the real and imaginary parts are negative, the angle lies in the **third quadrant**: \[ \theta = 180^\circ + 45^\circ = 225^\circ \] **Final Answer:** \[ Z_{1} + Z_{2} = 4.24 \angle 225^\circ \, \Omega \] --- ##### **3.2.2 \( \mathbf{Z_{1} - Z_{2}} \) and convert your answer into the polar form.** **Step 1: Calculate \( Z_{1} - Z_{2} \)** \[ Z_{1} - Z_{2} = (5 + j9) - (-8 - j12) = 5 + 8 + j(9 + 12) = 13 + j21 \, \Omega \] **Step 2: Convert to Polar Form** - **Magnitude (\( |Z| \)):** \[ |Z| = \sqrt{13^2 + 21^2} = \sqrt{169 + 441} = \sqrt{610} \approx 24.70 \, \Omega \] - **Phase Angle (\( \theta \)):** \[ \theta = \tan^{-1}\left(\frac{21}{13}\right) \approx \tan^{-1}(1.615) \approx 58^\circ \] Since both the real and imaginary parts are positive, the angle lies in the **first quadrant**. **Final Answer:** \[ Z_{1} - Z_{2} = 24.70 \angle 58^\circ \, \Omega \] --- ### **Summary of Results:** - **3.1 Five Detrimental Effects of Poor Power Factor:** 1. Increased Power Losses 2. Higher Electricity Bills 3. Reduced System Capacity 4. Overloading of Electrical Components 5. Voltage Drops and Poor Voltage Regulation - **3.2.1 \( Z_{1} + Z_{2} \):** \[ Z_{1} + Z_{2} = 4.24 \angle 225^\circ \, \Omega \] - **3.2.2 \( Z_{1} - Z_{2} \):** \[ Z_{1} - Z_{2} = 24.70 \angle 58^\circ \, \Omega \]