Fundamentals ofElectric circuits

In summary, the load in Fig. 11.44 draws 10 kVA at PF = 0.8 lagging and requires a reactive power of 8 kW. To achieve a PF of 0.9 lagging, a capacitor needs to be added, resulting in a reduction in reactive power. To find the capacitor value, the voltage across the source must be known, which can be found using the initial condition without the capacitor.
  • #1
Kokuson
4
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: 411
Physics news on Phys.org
  • #2
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.
 
  • #3


I am not able to provide a complete solution to this problem without more information. However, I can provide some guidance on how to approach the problem.

First, we need to understand the given information. The load in the circuit is drawing 10 kVA (kilovolt-amperes) at a power factor (PF) of 0.8 lagging. This means that the load is consuming a certain amount of reactive power, in addition to the active power. Reactive power is measured in units of volt-amperes reactive (VAR) and represents the power needed to create the magnetic field in inductive components in the circuit.

Next, we are given that the magnitude of the current, |IL|, is 40 A rms (root mean square). This means that the current is alternating at 40 A, and the rms value is the equivalent DC value that would produce the same amount of power in a resistive load.

To solve for the value of C, we need to use the equations for power in AC circuits. Specifically, we can use the equation P = |I| |V| cos@, where P is the active power, |I| is the magnitude of the current, |V| is the magnitude of the voltage, and @ is the phase angle between the current and voltage.

We know that the power factor is equal to cos@, so we can rearrange the equation to solve for @. Once we know @, we can use the equation @ = @V - @I to find the phase angle between the voltage and current.

Next, we can use the equation X = Xrms*(2)^0.5 to find the reactance of the circuit, where X is the reactance and Xrms is the rms value of the reactance. This will help us determine the value of the capacitor, C, needed to achieve a power factor of 0.9 lagging.

Finally, we can use the equation S = 0.5 * I * V (cos@ - j sin@) to calculate the complex power in the circuit, where S is the complex power, I is the current, V is the voltage, and j is the imaginary unit. This will help us verify that the circuit is operating at a power factor of 0.9 lagging.

Overall, the key to solving this problem is understanding the fundamentals of electric circuits and using the appropriate equations to solve for the unknown values
 

Related to Fundamentals ofElectric circuits

1. What is the purpose of studying fundamentals of electric circuits?

The purpose of studying fundamentals of electric circuits is to understand the principles and laws that govern the behavior of electricity, as well as how to analyze and design basic electrical circuits.

2. What are the basic components of an electric circuit?

The basic components of an electric circuit include a power source (such as a battery or generator), conductors (such as wires), resistors, capacitors, and inductors.

3. How do you calculate the voltage, current, and resistance in a circuit?

Using Ohm's law, voltage (V) is calculated by multiplying the current (I) by the resistance (R). Current is measured in amperes (A) and resistance in ohms (Ω). Resistance can also be calculated by dividing the voltage by the current.

4. What is the difference between series and parallel circuits?

In a series circuit, components are connected one after the other, so the same current flows through each component. In a parallel circuit, components are connected side by side, so the total current is divided among each component.

5. How can you analyze and solve complex electric circuits?

To analyze and solve complex electric circuits, you can use techniques such as Kirchhoff's laws, Thevenin's theorem, and Norton's theorem. These methods help to simplify the circuit and make it easier to calculate voltages and currents at different points.

Similar threads

  • Engineering and Comp Sci Homework Help
Replies
2
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
1
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
18
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
5
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
8
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
16
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
6
Views
3K
  • Engineering and Comp Sci Homework Help
Replies
4
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
30
Views
3K
Back
Top