Electrical Engineering - circuits - Reactive Networks

In summary: Vo being the Thevenin Voltage.It looks like you are going to need to review your text or course notes for more information on how to determine the Thevenin voltage.In summary, using Thevenin's theorem, the circuit contains a single voltage source and a series impedance. To calculate the Thevenin voltage, remove the load and find the voltage at the output terminals.
  • #36
I've re-done my working out for Vo considering that 1/j=-j, am I correct?

20160130_154921.jpg
 
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  • #37
Check the value you've used for the capacitor reactance. It should be 10/7, not 7/10. The capacitor impedance is ##-j\frac{10}{7}##.
 
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  • #38
Ah I understand now! Using -10/7 as the capacitor's impedance I got Vo= 2.6257 + j9.7993
 
  • #39
Yup. Much better. Be sure to use this corrected capacitor impedance in your Thevenin Impedance calculation, too.
 
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  • #40
For the Thevenin Impedance, do I need to take into account the resistor from the left side of the circuit? Or just the impedance from the capacitor and inductor?
 
  • #41
Frankboyle said:
For the Thevenin Impedance, do I need to take into account the resistor from the left side of the circuit? Or just the impedance from the capacitor and inductor?
If you suppress both sources, what does the resulting network look like? Is there any current path through the resistor?
 
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  • #42
If both sources are suppressed then there wouldn't be any current flowing through the resistor, so it shouldn't affect the Thevenin Impedance
 
  • #43
Frankboyle said:
If both sources are suppressed then there wouldn't be any current flowing through the resistor, so it shouldn't affect the Thevenin Impedance
Correct. Note also that you previously determined that the short circuit effectively divides the circuit into two isolated circuits. You can ignore the current source subcircuit entirely for any behavior or properties of the voltage source subcircuit, and vice versa.
 
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  • #44
The question says (Use peak values for the voltages and currents in your calculation)
So shouldn't you multiply 70/71 by the peak value (and not the phasor which was done) for V which is 10? and leave it as no real part? so 0 + j( 70/71 x 10 )

or why else does it clearly say use the peak values?
 
  • #45
You need to include the phases when doing calculations. The phasors you're working with are based on the peak voltages and currents (as opposed to rms). Your voltage supply comes with a built-in phase shift of 75°. Connecting an ideal voltage source to a circuit doesn't change its intrinsic phase. So you must incorporate that fact into the eventual Thevenin voltage that you find. That's why you use the phasor of the voltage source.

I can't think of any particular reason why they want you to use "peak phasors", other than perhaps that they're hoping to trip you up when you get to part (e) :wink: (hint, hint...).
 
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  • #46
Alright thanks,

And for nortons, is it right to assume that all the current in the circuit runs into the load:

since you have to short circuit the load to find the current for all the circuit's impedance
and then divide the voltage provided by this impedance to find the circuits current,

and since there's no element leading into the load, there's no current divider to solve, thus all the current runs into the load?

If I've said too much by all means remove my post :( but could you at least let me know if I am on the right track as I don't see the point creating a new thread when I am doing the exact same thing as the original poster
 
  • #47
@johnwillbert82 It's fine to tag along on this post to ask questions or provide help to the OP. Just don't provide complete solutions to his queries.

Regarding the Norton equivalent, you can start from scratch to determine it, applying the usual procedure of determining the short circuit current and so on. Or, which is more expedient, directly convert the Thevenin model you already have to a Norton model. Their relationship is very simple.

You are correct that shorting the output effectively removes the inductor from consideration. You still have to deal with the capacitor and the phase inherent to the voltage supply.
 
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  • #48
So can it simply be a source transformation from your thevenins? with the new value for I being I=V/Z with Z being from the previous answer? and V being from the voltage source
 
  • #49
johnwillbert82 said:
So can it simply be a source transformation from your thevenins? with the new value for I being I=V/Z with Z being from the previous answer? and V being from the voltage source
##I_N = V_{th} / Z_{th}##. The Thevenin model must produce the same short-circuit behavior as the Norton model and the original circuit.
 
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  • #50
@gneill also do we now take into account the current source and resistor on the left? since there is no longer a short circuit, as there will now be a current source there
 
  • #51
johnwillbert82 said:
@gneill also do we now take into account the current source and resistor on the left? since there is no longer a short circuit, as there will now be a current source there
No. The short still divides the two circuits.

Here's the situation with the Thevenin model in place:

upload_2016-1-30_13-31-11.png


Replace the Thevenin model with a Norton model. What do you get?
 
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  • #52
Thanks! Any idea for part D and E as I am stumped :(
 
  • #53
johnwillbert82 said:
Thanks! Any idea for part D and E as I am stumped :(
I'll need to see an attempt from the OP before I can share :smile:
 
  • #54
Sorry about the delay, only just had a chance to give part C a go!

I'm feeling a bit more sure about this part, but I've attached my work anyway just to be sure
20160131_195513.jpg
 
  • #55
Ah. You've mixed the Thevenin model with the original circuit there. You can't just drop the Thevenin voltage into the original circuit as it already expresses the effects of the other components.

You can, on the other hand, start with the entire Thevenin equivalent circuit and proceed from there. A Thevenin equivalent entirely replaces the original circuit and exhibits the same behavior. So you can convert the Thevenin model directly to a Norton equivalent.
 
  • #56
So would this be the correct conversion?

20160131_201623.jpg
 
  • #57
Yes. It's common practice to rename ##Z_{th}## to ##Z_N## for the Norton model, even though they have the same value.
 
  • #58
So as for the calculations, where I've previously used Zc, should I just use Zth to work out In?
 
  • #59
Frankboyle said:
So as for the calculations, where I've previously used Zc, should I just use Zth to work out In?
Yes. Once you've got the Thevenin equivalent you can throw out the original circuit and never look at it again :smile:
 
  • #60
I've re-done my calculations using Zth instead of Zc, does this look better?
20160131_214943_0.jpg
 
  • #61
Your Thevenin Impedance ##Z_T## looks very odd. Where did it come from ?
 
  • #62
I added the impedance of the capacitor and inductor, is this not the correct method?
 
  • #63
Frankboyle said:
I added the impedance of the capacitor and inductor, is this not the correct method?
We examined that before. How are those components connected to each other when the voltage supply is suppressed?
 
<h2>1. What is a reactive network in electrical engineering?</h2><p>A reactive network is a type of electrical circuit that contains components such as capacitors and inductors, which can store and release energy. These components cause the circuit to behave differently than a purely resistive circuit, resulting in a phenomenon known as reactance.</p><h2>2. How do you calculate the reactance of a reactive network?</h2><p>The reactance of a reactive network can be calculated using the formula X = 1/(2πfC) for capacitors and X = 2πfL for inductors, where X is the reactance in ohms, f is the frequency in hertz, C is the capacitance in farads, and L is the inductance in henrys.</p><h2>3. What is the difference between reactance and resistance?</h2><p>Resistance is a measure of the opposition to current flow in a circuit, while reactance is a measure of the opposition to the change in current flow caused by the presence of reactive components. In other words, resistance affects the magnitude of current, while reactance affects the phase of current.</p><h2>4. How do reactive networks affect power in a circuit?</h2><p>Reactive networks can cause a phase shift between voltage and current in a circuit, resulting in a difference between the apparent power and the real power. This can lead to power losses and decreased efficiency in the circuit.</p><h2>5. What are some common applications of reactive networks?</h2><p>Reactive networks are commonly used in electronic filters, power supplies, and AC power transmission systems. They are also essential in the design of radio frequency circuits and in the control of motor speed in industrial applications.</p>

1. What is a reactive network in electrical engineering?

A reactive network is a type of electrical circuit that contains components such as capacitors and inductors, which can store and release energy. These components cause the circuit to behave differently than a purely resistive circuit, resulting in a phenomenon known as reactance.

2. How do you calculate the reactance of a reactive network?

The reactance of a reactive network can be calculated using the formula X = 1/(2πfC) for capacitors and X = 2πfL for inductors, where X is the reactance in ohms, f is the frequency in hertz, C is the capacitance in farads, and L is the inductance in henrys.

3. What is the difference between reactance and resistance?

Resistance is a measure of the opposition to current flow in a circuit, while reactance is a measure of the opposition to the change in current flow caused by the presence of reactive components. In other words, resistance affects the magnitude of current, while reactance affects the phase of current.

4. How do reactive networks affect power in a circuit?

Reactive networks can cause a phase shift between voltage and current in a circuit, resulting in a difference between the apparent power and the real power. This can lead to power losses and decreased efficiency in the circuit.

5. What are some common applications of reactive networks?

Reactive networks are commonly used in electronic filters, power supplies, and AC power transmission systems. They are also essential in the design of radio frequency circuits and in the control of motor speed in industrial applications.

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