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Please convert the below content to rendered equations without any errors.Got it — let’s solve it **slowly and cleanly, step by step**, exactly how you’d do it in an exam. ? --- ## Given data * **Real power (load)** [ P = 80~\text{kW} ] * **Line voltage** [ V = 220~\text{V}_{\text{rms}}, \quad f = 60~\text{Hz} ] * **Initial power factor** [ \cos\phi_1 = 0.65 \quad (\text{lagging}) ] * **Capacitance added in parallel** [ C = 3000~\mu\text{F} = 3000 \times 10^{-6}~\text{F} ] --- ## Step 1: Find initial phase angle [ \phi_1 = \cos^{-1}(0.65) ] [ \phi_1 \approx 49.46^\circ ] --- ## Step 2: Find initial reactive power of the load [ Q_1 = P \tan \phi_1 ] [ Q_1 = 80 \times \tan(49.46^\circ) ] [ Q_1 \approx 80 \times 1.17 = 93.6~\text{kVAR} ] ? This is **inductive reactive power (lagging)**. --- ## Step 3: Find capacitive reactive power added Angular frequency: [ \omega = 2\pi f = 2\pi \times 60 = 377~\text{rad/s} ] Capacitive reactive power: [ Q_C = V^2 \omega C ] Substitute values: [ Q_C = (220)^2 \times 377 \times (3000 \times 10^{-6}) ] [ Q_C = 48{,}400 \times 377 \times 0.003 ] [ Q_C \approx 54.7~\text{kVAR} ] ? This is **leading (capacitive)** reactive power. --- ## Step 4: Find new net reactive power [ Q_{\text{new}} = Q_1 - Q_C ] [ Q_{\text{new}} = 93.6 - 54.7 = 38.9~\text{kVAR} ] Still **lagging**, but much smaller. --- ## Step 5: Find new apparent power [ S_{\text{new}} = \sqrt{P^2 + Q_{\text{new}}^2} ] [ S_{\text{new}} = \sqrt{(80)^2 + (38.9)^2} ] [ S_{\text{new}} \approx \sqrt{6400 + 1513} \approx 89.0~\text{kVA} ] --- ## Step 6: Find new power factor [ \text{PF}*{\text{new}} = \frac{P}{S*{\text{new}}} ] [ \text{PF}_{\text{new}} = \frac{80}{89.0} ] [ \boxed{\text{PF}_{\text{new}} \approx 0.90 ; \text{lagging}} ] --- ## ✅ Final Answer [ \boxed{\text{New power factor} \approx 0.90 ; \text{lagging}} ] --- ### ? Key insight (important for exams) * Capacitors **do not change real power** * They **reduce reactive power** * Power factor improves because apparent power decreases If you want, I can: * Show a **phasor diagram** * Solve this assuming **3-phase system** * Tell you **how much capacitance is needed for unity PF** Just tell me ?

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Please convert the below content to rendered equations without any errors.Got it — let’s solve it **slowly and cleanly, step by step**, exactly how you’d do it in an exam. ? --- ## Given data * **Real power (load)** [ P = 80~\text{kW} ] * **Line voltage** [ V = 220~\text{V}_{\text{rms}}, \quad f = 60~\text{Hz} ] * **Initial power factor** [ \cos\phi_1 = 0.65 \quad (\text{lagging}) ] * **Capacitance added in parallel** [ C = 3000~\mu\text{F} = 3000 \times 10^{-6}~\text{F} ] --- ## Step 1: Find initial phase angle [ \phi_1 = \cos^{-1}(0.65) ] [ \phi_1 \approx 49.46^\circ ] --- ## Step 2: Find initial reactive power of the load [ Q_1 = P \tan \phi_1 ] [ Q_1 = 80 \times \tan(49.46^\circ) ] [ Q_1 \approx 80 \times 1.17 = 93.6~\text{kVAR} ] ? This is **inductive reactive power (lagging)**. --- ## Step 3: Find capacitive reactive power added Angular frequency: [ \omega = 2\pi f = 2\pi \times 60 = 377~\text{rad/s} ] Capacitive reactive power: [ Q_C = V^2 \omega C ] Substitute values: [ Q_C = (220)^2 \times 377 \times (3000 \times 10^{-6}) ] [ Q_C = 48{,}400 \times 377 \times 0.003 ] [ Q_C \approx 54.7~\text{kVAR} ] ? This is **leading (capacitive)** reactive power. --- ## Step 4: Find new net reactive power [ Q_{\text{new}} = Q_1 - Q_C ] [ Q_{\text{new}} = 93.6 - 54.7 = 38.9~\text{kVAR} ] Still **lagging**, but much smaller. --- ## Step 5: Find new apparent power [ S_{\text{new}} = \sqrt{P^2 + Q_{\text{new}}^2} ] [ S_{\text{new}} = \sqrt{(80)^2 + (38.9)^2} ] [ S_{\text{new}} \approx \sqrt{6400 + 1513} \approx 89.0~\text{kVA} ] --- ## Step 6: Find new power factor [ \text{PF}*{\text{new}} = \frac{P}{S*{\text{new}}} ] [ \text{PF}_{\text{new}} = \frac{80}{89.0} ] [ \boxed{\text{PF}_{\text{new}} \approx 0.90 ; \text{lagging}} ] --- ## ✅ Final Answer [ \boxed{\text{New power factor} \approx 0.90 ; \text{lagging}} ] --- ### ? Key insight (important for exams) * Capacitors **do not change real power** * They **reduce reactive power** * Power factor improves because apparent power decreases If you want, I can: * Show a **phasor diagram** * Solve this assuming **3-phase system** * Tell you **how much capacitance is needed for unity PF** Just tell me ?

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Asked by: jack
Created at: 2026-02-27 07:53:47

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# Step-by-Step Solution: Power Factor Correction with Capacitor --- ## Given Data - **Real power (load):** \[ P = 80~\text{kW} \] - **Line voltage:** \[ V = 220~\text{V}_{\text{rms}}, \quad f = 60~\text{Hz} \] - **Initial power factor:** \[ \cos\phi_1 = .65 \quad (\text{lagging}) \] - **Capacitance added in parallel:** \[ C = 300~\mu\text{F} = 300 \times 10^{-6}~\text{F} \] --- ## Step 1: Find Initial Phase Angle \[ \phi_1 = \cos^{-1}(.65) \] \[ \phi_1 \approx 49.46^\circ \] --- ## Step 2: Find Initial Reactive Power of the Load \[ Q_1 = P \tan \phi_1 \] \[ Q_1 = 80 \times \tan(49.46^\circ) \] \[ Q_1 \approx 80 \times 1.17 = 93.6~\text{kVAR} \] (This is **inductive reactive power, lagging**.) --- ## Step 3: Find Capacitive Reactive Power Added Angular frequency: \[ \omega = 2\pi f = 2\pi \times 60 = 377~\text{rad/s} \] Capacitive reactive power: \[ Q_C = V^2 \omega C \] Substitute values: \[ Q_C = (220)^2 \times 377 \times (300 \times 10^{-6}) \] \[ Q_C = 48{,}400 \times 377 \times .003 \] \[ Q_C \approx 54.7~\text{kVAR} \] (**This is leading (capacitive) reactive power.**) --- ## Step 4: Find New Net Reactive Power \[ Q_{\text{new}} = Q_1 - Q_C \] \[ Q_{\text{new}} = 93.6 - 54.7 = 38.9~\text{kVAR} \] (**Still lagging, but much smaller.**) --- ## Step 5: Find New Apparent Power \[ S_{\text{new}} = \sqrt{P^2 + Q_{\text{new}}^2} \] \[ S_{\text{new}} = \sqrt{(80)^2 + (38.9)^2} \] \[ S_{\text{new}} \approx \sqrt{640 + 1513} \approx 89.~\text{kVA} \] --- ## Step 6: Find New Power Factor \[ \text{PF}_{\text{new}} = \frac{P}{S_{\text{new}}} \] \[ \text{PF}_{\text{new}} = \frac{80}{89.} \] \[ \boxed{\text{PF}_{\text{new}} \approx .90 \quad \text{(lagging)}} \] --- ## Final Answer \[ \boxed{\text{New power factor} \approx .90 \quad \text{(lagging)}} \] --- ## Key Insights - **Capacitors do not change real power** - **They reduce reactive power** - **Power factor improves because apparent power decreases** --- If you want to see a **phasor diagram**, a **3-phase version**, or the **capacitance needed for unity PF**, just ask.

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