Problem about Sinusoidal Circuits

[BRAWNY]

hey everybody
how are you ?
hope everyone's fine
well , i have this problem related to electrical circuits

Homework Statement

A group of small appliances on a 60 Hz system requires 20kVA at 0.85 pf
lagging when operated at 120 V (rms). The impedance of the feeder
supplying the appliances is 0.01 + j0.08Ω. The voltage at the load end of
the feeder is 120 V.

Homework Equations

1. What is the rms magnitude of the voltage at the source end of the feeder?
2. What is the average power loss in the feeder?
3. What size capacitor (in microfarads) at the load end of the feeder is
needed to improve the load power factor to unity?
4. After the capacitor is installed, what is the rms magnitude of the voltage
at the source end of the feeder if the load voltage is maintainedDW V?
5. What is the average power loss in the feeder for part (d)?

The Attempt at a Solution

well , since the problem gives us the complex power (20000 VA ) and the Voltage , i can get the conjugate current which equals to 166,6 and the phase angle i can get it from the power factor (0.8) .. that is what i know ..
please if you can help me out here i will be appreciated

 

[Antiphon]

Draw a circuit which includes the feeder impedance since the power factor and voltage are given at this point, not at the ideal generator.

If you don't see why this is needed, let me say that the ideal voltage source is slightly higher than 120 volts. If it's not obvious why, then you need the schematic.

 

[BRAWNY]

i still have problems how to solve it

any help
i have to submit it tomorrow >> please help me out

 

[gneill]

It would be nice to see this thread completed, even if the homework is already past due.

 

[DeeNos]

Hi there,

I'm glad to hear that you are interested in learning about sinusoidal circuits! I can provide you with some guidance to help you solve this problem.

Firstly, let's break down the problem into smaller parts. We have a group of small appliances that require 20kVA of power at a power factor of 0.85 lagging. This means that the load is using more reactive power (VAR) than real power (W). The feeder supplying the appliances has an impedance of 0.01 + j0.08Ω. This means that it has both a resistance and a reactance. The voltage at the load end of the feeder is 120 V.

Now, let's look at the equations that we can use to solve this problem:

1. To find the rms magnitude of the voltage at the source end of the feeder, we can use the equation V = IZ, where V is the voltage, I is the current, and Z is the impedance. Since we know the voltage and the impedance, we can solve for the current (I) and then use that to find the rms magnitude of the voltage at the source end.

2. The average power loss in the feeder can be calculated using the equation P = I^2R, where P is the power loss, I is the current, and R is the resistance of the feeder. We can use the current that we found in the previous step and the resistance of the feeder (0.01Ω) to solve for the power loss.

3. To improve the power factor to unity, we need to add a capacitor to the load end of the feeder. The size of the capacitor can be calculated using the equation C = Q/V, where C is the capacitance, Q is the reactive power (VAR) that needs to be compensated, and V is the voltage. Since we know the reactive power (VAR) and the voltage, we can solve for the capacitance needed.

4. After the capacitor is installed, the voltage at the load end of the feeder will be maintained at 120 V. To find the rms magnitude of the voltage at the source end, we can use the equation V = IZ, where V is the voltage, I is the current, and Z is the impedance. However, the impedance will now be different since the power factor has been improved to unity. We can use the new impedance and the current that