Fundamentals ofElectric circuits

  • Context: Engineering 
  • Thread starter Thread starter Kokuson
  • Start date Start date
  • Tags Tags
    Circuits Fundamentals
Click For Summary
SUMMARY

The discussion focuses on calculating the required capacitance to improve the power factor (PF) of an electrical load from 0.8 lagging to 0.9 lagging. The load draws 10 kVA with a real power of 8 kW. The solution involves determining the reduction in reactive power needed to achieve the desired PF and using the voltage across the source to find the capacitance value. Key equations include the relationship between apparent power, real power, and power factor, as well as the calculation of reactive power.

PREREQUISITES
  • Understanding of power factor and its significance in electrical circuits
  • Familiarity with complex power equations, including S = P + jQ
  • Knowledge of reactive power and its role in power factor correction
  • Basic skills in circuit analysis and voltage calculations
NEXT STEPS
  • Study the concept of power factor correction using capacitors in AC circuits
  • Learn how to calculate reactive power and its impact on overall power consumption
  • Explore the use of phasor diagrams to visualize current and voltage relationships
  • Investigate the effects of different power factor values on electrical efficiency
USEFUL FOR

Electrical engineering students, professionals involved in circuit design, and anyone interested in improving energy efficiency through power factor correction techniques.

Kokuson
Messages
4
Reaction score
0

Homework Statement



The load in Fig. 11.44 draws 10 kVA at PF =0.8
lagging. If |IL |=40 A rms, what must be the value of C
to cause the source to operate at PF =0.9 lagging?

Homework Equations



P(apparent power VA)= |I| |V| cos@
@ = @V - @I
cos@= p.f
X= Xrms*(2)^0.5
S(complex power)= 0.5 * I * V (cos@ - j sin@)

The Attempt at a Solution


I just face a difficulty in knwoing the face of IL
I will be thankful if you solved the wole problem.
 

Attachments

  • Ele204.gif
    Ele204.gif
    11.8 KB · Views: 468
Physics news on Phys.org
First, there is no capacitor. And, you have 10 kVA at 0.8 so 8 kW.

You add a capacitor and you get 10 kVA at 0.9 and 8KW

Find the reduction in the reactive power required to make the pf equal to 0.9. To find the capacitor value now you have its reactive power and you need voltage which is equal to the voltage across the source.

You can find the voltage across the source using the initial condition when there is no capacitor.
 

Similar threads

  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 30 ·
2
Replies
30
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K