Finding Power Dissipation in Three Series Elements

In summary: PF) can be found using the formula PF = cosθ, where θ is the angle between the total voltage and current. From part b), we know that this angle is 20.1 degrees. Therefore, the PF is equal to cos(20.1) = 0.94. This means that the overall power factor for this circuit is 0.94, which is fairly close to 1 and indicates a relatively efficient use of power. d) To find the total power, we can use the formula P = VIcosθ, where V is the RMS voltage, I is the RMS current, and θ is the angle between them. From parts a) and b), we know that the
  • #1
suspenc3
402
0

Homework Statement



Three series elements have a.c. voltages across them of [tex]25cos(2 \pi 60t + 30') , 15cos(2 \pi 60t + 60') , 30cos(2 \pi 60t -45') [/tex] Volts respectively. The current flowing is [tex]5.2cos(2 \pi 60t + 25')A[/tex]

Using phasor diagram find

a)total voltage
b)phase angle between total voltage and current
c)overall power factor
d)total power and power dissipated by each element


The Attempt at a Solution



I found part a to be [tex]50.5cos(2 \pi 60t + 4.9')[/tex]

and part be, the angle is 20.1 degrees with the current lagging behind.

I found the PF to be 0.94, what about d)power diss...i found the total power to be 123W
 
Last edited:
Physics news on Phys.org
  • #2


Hi there!

Thank you for your post. It seems like you have made a good start in solving this problem. Let's go through each part of the problem and see if we can come up with a complete solution.

a) To find the total voltage, we need to add the voltages of each element using phasor addition. Remember, phasors are vectors that represent the magnitude and phase of a sinusoidal quantity. We can draw a phasor diagram to help visualize this addition.

First, let's convert the given voltages into phasor form. The three series elements have voltages of 25cos(2 \pi 60t + 30') , 15cos(2 \pi 60t + 60') , 30cos(2 \pi 60t -45') Volts respectively. Using Euler's formula, we can convert these to phasor form as follows:

25cos(2 \pi 60t + 30') = 25e^(j(2 \pi 60t + 30')/180)

15cos(2 \pi 60t + 60') = 15e^(j(2 \pi 60t + 60')/180)

30cos(2 \pi 60t -45') = 30e^(j(2 \pi 60t -45')/180)

Now, let's plot these phasors on a phasor diagram. Since the voltages are all in phase with each other (they all have the same frequency and are in the form of cosines), we can simply add them together as vectors.

After adding the phasors, we get a total voltage of 50.5e^(j(2 \pi 60t + 4.9')/180) Volts. This is equivalent to 50.5cos(2 \pi 60t + 4.9') Volts in cosine form.

b) To find the phase angle between the total voltage and current, we can use the phasor diagram we just created. The angle between the total voltage and current is the same as the angle between the phasors representing these quantities. From the diagram, we can see that this angle is 20.1 degrees, which means that the total voltage is lagging behind the current by 20.1 degrees.

c) The
 
  • #3


I would first like to commend you on your efforts in finding the total voltage, phase angle, and power factor. Your calculations seem to be accurate and well thought out.

To find the total power and power dissipated by each element, we can use the formula P = VIcosθ, where P is power, V is voltage, I is current, and θ is the phase angle between voltage and current.

Using this formula, we can calculate the total power to be 128.4 watts (P = 50.5 x 5.2 x cos(20.1) = 128.4). This means that the total power dissipated by the three series elements is 128.4 watts.

To find the power dissipated by each element, we can use the same formula and substitute the voltage and current values for each element. The power dissipated by the first element is 54.1 watts (P = 25 x 5.2 x cos(55.1) = 54.1), the second element is 30.9 watts (P = 15 x 5.2 x cos(55.1) = 30.9), and the third element is 43.4 watts (P = 30 x 5.2 x cos(-65.1) = 43.4).

In summary, the total power dissipated by the three series elements is 128.4 watts, with 54.1 watts being dissipated by the first element, 30.9 watts by the second element, and 43.4 watts by the third element.

I hope this helps clarify the power dissipation in this circuit. Keep up the good work in your scientific endeavors!
 

What is power dissipation?

Power dissipation is the rate at which energy is converted into heat or other forms of energy by an electronic circuit. It is typically measured in watts (W) and is a measure of the amount of power that is lost or expended in a given circuit component.

What are series elements?

Series elements are electronic components that are connected in a series circuit, meaning that they are connected end-to-end with no branching. In a series circuit, the current remains constant throughout the circuit, but the voltage may vary across each element.

How do I calculate power dissipation in three series elements?

To calculate power dissipation in three series elements, you will need to know the voltage and current values for each element. You can then use the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amps. Simply calculate the power dissipation for each element and add them together to get the total power dissipation.

Why is it important to calculate power dissipation in series elements?

Calculating power dissipation in series elements is important for several reasons. It helps us understand how much power is being consumed or lost in a circuit, which can impact the efficiency and performance of the circuit. It also helps us determine if any components are at risk of overheating, which can lead to malfunction or damage.

Are there any limitations to calculating power dissipation in series elements?

Yes, there are some limitations to calculating power dissipation in series elements. This method assumes that the circuit is operating under normal conditions and does not take into account any potential variations or fluctuations in voltage and current. It also does not account for any external factors that may affect power dissipation, such as temperature or environmental conditions.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
7K
  • Introductory Physics Homework Help
Replies
28
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
3K
Replies
2
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
2K
Back
Top