Engineering Electrical Engineering - circuits - Reactive Networks

[Frankboyle]

Here's my working out for the voltage divider

 

[gneill]

Okay. Note that the impedance of a capacitor is negative. 1/j becomes -j when you move it to the numerator.

 

[johnwillbert82]

[moderator note: partial solution removed --- Please do not solve the Original Posters problems for them. Hints and suggestions are fine, but they must do the work themselves.]

Does the fact the impedances only have an imaginary part make a difference when working out VT using the voltage divider? Do you still multiply both parts of the voltage by Z2/(Z1 + Z2) which is what I did, or would you multiply the real part by 0 making VT = 0 + j9.7993

 

[gneill]

Does the fact the impedances only have an imaginary part make a difference when working out VT using the voltage divider? Do you still multiply both parts of the voltage by Z2/(Z1 + Z2) which is what I did, or would you multiply the real part by 0 making VT = 0 + j9.7993

Impedances are complex values and you use the entire value, both real and complex parts, when determining voltages, currents, or net impedance. The time to worry about real and imaginary parts separately is when you're dealing with power dissipation.

 

[johnwillbert82]

Right I see,

sorry for posting, I was just checking that I did it correct as I also have the same question as him... ill just make my own thread

 

[Frankboyle]

I've re-done my working out for Vo considering that 1/j=-j, am I correct?

 

[gneill]

Check the value you've used for the capacitor reactance. It should be 10/7, not 7/10. The capacitor impedance is $-j\frac{10}{7}$.

 

[Frankboyle]

Ah I understand now! Using -10/7 as the capacitor's impedance I got Vo= 2.6257 + j9.7993

 

[gneill]

Yup. Much better. Be sure to use this corrected capacitor impedance in your Thevenin Impedance calculation, too.

 

[Frankboyle]

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?

 

[gneill]

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?

 

[Frankboyle]

If both sources are suppressed then there wouldn't be any current flowing through the resistor, so it shouldn't affect the Thevenin Impedance

 

[gneill]

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.

 

[johnwillbert82]

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?

 

[gneill]

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) (hint, hint...).

 

[johnwillbert82]

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

 

[gneill]

@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.

 

[johnwillbert82]

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

 

[gneill]

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.

 

[johnwillbert82]

@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

 

[gneill]

@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:

Replace the Thevenin model with a Norton model. What do you get?

 

[johnwillbert82]

Thanks! Any idea for part D and E as I am stumped :(

 

[gneill]

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

 

[Frankboyle]

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

 

[gneill]

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.

 

[Frankboyle]

So would this be the correct conversion?

 

[gneill]

Yes. It's common practice to rename $Z_{th}$ to $Z_N$ for the Norton model, even though they have the same value.

 

[Frankboyle]

So as for the calculations, where I've previously used Zc, should I just use Zth to work out In?

 

[gneill]

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

 

[Frankboyle]

I've re-done my calculations using Zth instead of Zc, does this look better?