Solve RL Circuit Power Problem:V source 140V, 33A load, X=5R, Vload 115V, 3kW

In summary: W The reactive power at the load (Ql) is equal to the product of the voltage at the load and the current through the load times the sine of the angle between the voltage and current. Since the angle between the voltage and current is 90 degrees, Ql = VL x IL x sin(90) = 115V x 33A x 1 = 0.77 kW Now that we have calculated the active and reactive power at the load, we can determine the percentage of active power lost as feeder heat losses. The total power supplied by the source (Ps) is 3kW. Therefore, the power lost as feeder heat losses is equal to the difference between the total power supplied and the
  • #1
alinec613
1
0
An AC source of 140V feeds 33A to a load,
though a cable represented by an unknown
resistance R in series with an unknown
inductive reactance X. The manufacturer
of that cable says, in the catalog, that
X=5R. The voltage at the load is 115V.
The power supplied by the source is measured
as 3kW. What are the active and reactive power
at the load? What percentage of the active
power delivered by the source is lost as
feeder heat losses?


This is what I have done so far

http://answerboard.cramster.com/Answer-Board/Image/200711832026333008880273800009062.jpg

Is is parallel to Vr ; angle btwn Vr and Vx = 90 degrees
VL indicates V load, Vs = V source, Vr = V resistor, Vx = V inductor

The prof gave us a hint saying that "nothing stops you from extending the lines"
but I stilll can't get all the angles and lengths.

Someone please help. I really need to know how to solve this question.
 
Last edited:
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  • #2
Thanks.Answer:The active power at the load is 2.51 kW and the reactive power at the load is 0.77 kW. The percentage of active power lost as feeder heat losses is 20%. To solve this problem, we need to determine the resistance and inductive reactance of the cable. Since X=5R, we can set up the following equation:5R = X = ωL = 2πfL Where ω is the angular frequency and f is the frequency of the AC supply. We know that the voltage at the load (VL) is 115V and the voltage at the source (Vs) is 140V. Therefore, the total voltage drop across the cable (Vc) is 25V. Since the cable has a resistance (R) in series with an inductive reactance (X), the voltage drop across each element can be written as:Vc = Vr + Vx Where Vr is the voltage drop across the resistance and Vx is the voltage drop across the inductive reactance. We can use Ohm's Law to calculate the voltage drop across the resistance as follows:Vr = IR Where I is the current through the cable. We know that the current through the cable is 33A and the resistance is unknown. Therefore, we can rearrange the equation to calculate the resistance:R = Vr/I Substituting the values, we get: R = 25V/33A = 0.75 Ω Now that we have determined the resistance of the cable, we can calculate the inductive reactance using the equation X = 5R. X = 5R = 5 x 0.75Ω = 3.75 Ω Now that we know the resistance and inductive reactance of the cable, we can calculate the active and reactive power at the load. The active power at the load (Pl) is equal to the product of the voltage at the load and the current through the load. Therefore, Pl = VL x IL = 115V x 33A =
 
  • #3


To solve this problem, we can use the equations for power in an RL circuit:

Active power (P) = Vload * Iload * cos(theta)
Reactive power (Q) = Vload * Iload * sin(theta)

First, we need to find the values of R and X. We know that X = 5R, so we can set up the following equation:

X = 5R
X/R = 5
tan(theta) = 5

Using a calculator, we can find that theta is approximately 78.69 degrees.

Next, we can use the voltage and current values to find the values of R and X. We know that Vload = 115V, Iload = 33A, and theta = 78.69 degrees. Using the equations for voltage and current in an RL circuit, we can set up the following equations:

Vload = Vr + Vx
Iload = (Vs - Vr)/R

Solving for Vr and R, we get:

Vr = Vload - Vx
R = (Vs - Vr)/Iload

Plugging in the values, we get:

Vr = 115V - 115V * tan(78.69 degrees) = 0V
R = (140V - 0V)/33A = 4.24 ohms

Now, we can find the values of P and Q using the equations mentioned earlier:

P = 115V * 33A * cos(78.69 degrees) = 3kW
Q = 115V * 33A * sin(78.69 degrees) = 2.82kVAR

Finally, to find the percentage of active power lost as feeder heat losses, we can use the equation:

% active power loss = (Psource - Pload)/Psource * 100%

Since Psource = 3kW and Pload = 3kW, the percentage of active power lost is 0%. This means that all of the power supplied by the source is used by the load, and there are no losses due to the cable.
 

1. How do I calculate the power in an RL circuit?

To calculate the power in an RL circuit, you can use the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amps. In this problem, we are given the voltage source, load current, and load voltage, so we can plug those values into the formula to solve for power.

2. What is the significance of X in an RL circuit?

X represents the reactance in an RL circuit, which is the opposition to the flow of alternating current. It is measured in ohms and is dependent on the inductance and frequency of the circuit. In this problem, X=5R means that the reactance is 5 times the resistance.

3. How does the power output of an RL circuit change with different load values?

The power output of an RL circuit is dependent on the load current and voltage. As the load resistance changes, the load current and voltage will also change, resulting in a different power output. In this problem, the load resistance is 3kW, which is a high resistance and will result in a lower power output compared to a lower load resistance.

4. What is the relationship between voltage and power in an RL circuit?

In an RL circuit, power is directly proportional to voltage. This means that as the voltage increases, the power output will also increase. In this problem, the voltage source is 140V, while the load voltage is 115V. This means that the power output may be lower than the maximum potential power output of the circuit.

5. How does the inductance of the circuit affect the power output?

The inductance of an RL circuit affects the power output by increasing the reactance and therefore decreasing the current flow. This results in a lower power output. In this problem, the inductance is not given, but it is important to note that a higher inductance would result in a lower power output.

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