Engineering Complex current calculation in Parallel R L Circuit

[gl0ck]

Homework Statement

I need to find the complex current in this circuit
http://imageshack.us/photo/my-images/51/figure1j.jpg/

Homework Equations

I calculated the inductor's impedance and found that impedance = 12.57 Ohms
so I used it in the equation for the Current.
I need to write the answer in terms of a + bj (complex numbers)

The Attempt at a Solution

I tried to find the current with the Ohm's law I = V/Z where Zt = 1/1/Zr+1/Zl
I am not sure is that is the right way to calculate this.
Thanks

 

[gneill]

Homework Statement

I need to find the complex current in this circuit
http://imageshack.us/photo/my-images/51/figure1j.jpg

Homework Equations

I calculated the inductor's impedance and found that impedance = 12.57 Ohms
so I used it in the equation for the Current.
I need to write the answer in terms of a + bj (complex numbers)

The Attempt at a Solution

I tried to find the current with the Ohm's law I = V/Z where Zt = 1/1/Zr+1/Zl
I am not sure is that is the right way to calculate this.
Thanks

Which current are you trying to find? There are two parallel branches: one consisting of a capacitor of unknown value, and one consisting of the SERIES connection of an inductor and a resistor. So it looks like there are three different currents to choose from.

 

[gl0ck]

Take a look of the question it is said the complex current of the circuit

http://imageshack.us/photo/my-images/696/figure1u.png/

 

[Zarathuztra]

I couldn't see the circuit, but as for bringing it to terms of a + bj... Impedance of the resistor is just R, the inductor is Z = jωL so the admittance is Y = -j1/L (same as 1/jωL but I keep complex numbers in the numerator).

any reason I couldn't see the circuit?

 

[gneill]

From the problem statement in your second image (yes, for some reason your images are not being displayed in your posts; I hit "quote" to obtain the full text of the post and copied the URL from out of the IMG tags) you need to find the impedance of the load algebraically to start with. That means ignoring the numerical component values to begin with and deriving an expression for the total impedance symbolically.

So using L, R, and C for the component values, can you write an expression for the impedance?

 

[gl0ck]

So Ztotal must be equal to 1/(1/Zr+1/Zl) + Zc?

 

[aralbrec]

The resistor and inductor are not in parallel.

The capacitor is in parallel with the series resistor + inductor.

 

[gneill]

So Ztotal must be equal to 1/(1/Zr+1/Zl) + Zc?

No, the resistor is in series with the inductor (so their impedances add) while the capacitor is in parallel with their sum.

It occurs to me that it might be simpler to write the currents of the two branches separately (since they are in parallel they share the same potential difference of E = 220V), then sum the results. This will tame the algebra which might otherwise get quite hairy.

 

[gl0ck]

Oh, I see the stupid mistake .. So the total impedance of the Resistor and Inductor is 8+12.57=20.57Ohms
But how to get rid of the Capacitor.. because they ask me about the capacitor in the second part of the question.
It seems that the final equation should be I= V/1/(1+Zc+1/Zr+l)

 

[gneill]

Oh, I see the stupid mistake .. So the total impedance of the Resistor and Inductor is 8+12.57=20.57Ohms

Remember that impedances are complex values. The impedance of the inductor is 12.57j Ohms, and doesn't add to the real component of the resistor impedance.

But how to get rid of the Capacitor.. because they ask me about the capacitor in the second part of the question.
It seems that the final equation should be I= V/1/(1+Zc+1/Zr+l)

Forget the numbers for now. Do the math symbolically. If the capacitor has value "C", and the power supply has voltage "E", what's the (complex) current through the capacitor branch? Do the same for the Jacuzzi branch, using R and L for the component values.

Separate the currents into their real and imaginary components and add the 'like' components together to arrive at the total current.

 

[aralbrec]

Oh, I see the stupid mistake .. So the total impedance of the Resistor and Inductor is 8+12.57=20.57Ohms

You can't add them together like that. R is a *real* resistance while L is an *imaginary* reactance. The series impedance is 8 + j12.57 with the imaginary reactance being separated from the real resistance by the 'j'.

This has to do with the current and voltage relationships that these devices impose across them. A resistor is governed by V=IR, where the current and voltage are always in phase (there is no time delay between the two waveforms -- if you place 10 volts across a 1 ohm resistor, 1 amp flows through it immediately, not 10 seconds from now). An inductor is governed (in the steady state**) by the same equation: V = I (jwL) but the resistance is complex. This is because a time delay is introduced in the V/I relationship and this is summarized by the phase displacement introduced by the complex 'j'. If you apply V volts across an inductor, the current through the inductor does *not* change immediately but causes the current to change gradually according to di/dt = v/L. This time delay is the phase seen in complex impedances.

But how to get rid of the Capacitor.. because they ask me about the capacitor in the second part of the question.
It seems that the final equation should be I= V/1/(1+Zc+1/Zr+l)

I don't know what they are asking about the question but the I expression you have looks to be missing some brackets.

The impedance of the series R/L is ZL = R + jwL and the impedance of the cap is Zc. So you need the parallel combination of those two to find the impedance of the circuit seen by the source.

BTW -- it is much easier not to make algebraic errors and to sanity check your work if you wait to plug in actual values until the very end. It will save you much hassle :)

 

[gl0ck]

The impedance of the series R/L is ZL = R + jwL and the impedance of the cap is Zc. So you need the parallel combination of those two to find the impedance of the circuit seen by the source.

The thing I come up with is the following I=V/1/(1/Zc+1/(R+R+jwL)) isn't the whole jwL = Zl or it is Zl=R+jwl

and I still don't get how to get rid of Zc becuase otherwise we will have 2 unknowns..

 

[aralbrec]

The thing I come up with is the following I=V/1/(1/Zc+1/(R+R+jwL))

That will be the total current delivered by the source (see the equation I marked with ******* below which is equivalent for the impedance calculation)

but you have an extra R there. The inductor impedance by itself is jwL. The resistor has impedance R. The sum is R+jwL

isnt the whole jwL = Zl or it is Zl=R+jwl

Maybe I chose an unfortunate symbol. The impedance of the inductor alone is jwL. The impedance of the resistor in series with the inductor (which I called ZL or Z load) is R+jwL

and I still don't get how to get rid of Zc becuase otherwise we will have 2 unknowns..

The voltage E appears across the series R/L so the current through R/L can be found independently of the capacitor and is E/(R+jwL)

The total current supplied by the source is the sum of the current through the capacitor and the current through the series R/L. I(total) = E/(R+jwL) + EjwC.

You could also find the total current by finding the total impedance seen by the source. This is Z(total) = (R+jwL) || (1/jwC) ********* and I(total) = E / Z(total)How you go about answering the question will depend on what information is given and what is asked. I can only see the circuit diagram though :)

 

[gl0ck]

So, for the first part a) the total current doesn't have anything to do with Zc..
so its I=E/R+jwL
and doing the nasty algebra I came to this E(R-jwL)/R^2-jwL^2
220(8+j12.57)/166,0049
can someone tell me how to remove the j from the above equation since i know that Zr+l = 8+j12.57
and j^2 =- 1

 

[gneill]

So, for the first part a) the total current doesn't have anything to do with Zc..

Actually, it does have something to do with Zc. You want to find an expression for the total current which includes the capacitor so that you can apply the hint given for part b.
However, you can proceed by finding the currents for the two branches separately, then add the results to find the total.

so its I=E/R+jwL

You should make liberal use of parentheses to make the interpretation of your equations clear. Presumably you meant,

I = E/(R + jωL)

and doing the nasty algebra I came to this E(R-jwL)/R^2-jwL^2

by which presumably you mean:

I = E(R - jωL)/(R² - (jωL)²)

can someone tell me how to remove the j from the above equation since i know that Zr+l = 8+j12.57
and j^2 =- 1

The j in the denominator goes away when you carry out the squaring of the (jωL) term. The j in the numerator stays as is.

This value for I is the current in the Jacuzzi branch. Now, what's the current in the capacitor branch?

 

[gl0ck]

so we have (ER - jEwL)/(R^2 + wL^2) and this gives us (220*8+j220*12.57)/166,0049

I think in the second part of the question which is Now that you have the complex current, what value for the shunt capacitor will minimise
the total current? Hint: Set the imaginary part to 0

we have to find the Zc?

 

[gneill]

so we have (ER - jEwL)/(R^2 + (wL)^2) and this gives us (220*8+j220*12.57)/166,0049

Note the added parentheses in order to make the mathematical operations clear.

You don't need any numbers yet! Just the symbols. Remember that the problem says you want to arrive at a formula you can apply to Bob's friend George's similar (but not the same) Jacuzzi. For that you need a formula containing symbols only.

I think in the second part of the question which is Now that you have the complex current, what value for the shunt capacitor will minimise
the total current? Hint: Set the imaginary part to 0.

we have to find the Zc?

You need to find the expression for the current WHICH INCLUDES THE CAPACITOR before you can do part b.

You have an expression for current of the Jacuzzi branch. Now write an expression for the current in the capacitor branch.

 

[gl0ck]

Zt=(RjwC)/(R+jwC-w^2LCR) ?

 

[gneill]

Zt=(RjwC)/(R+jwC-w^2LCR) ?

That doesn't look right. But why are you going back to the impedance? Just write the expression for the current through the capacitor; Since the two branches are in parallel, you can simply add the two currents to find the total.

 

[gl0ck]

isnt it when they are in parallel 1/Zt = 1/Zc + 1/Zr+l ?
so the current through cap will be I = V/Zc?

 

[gneill]

isnt it when they are in parallel 1/Zt = 1/Zc + 1/Zr+l ?

I don't understand your last term. What's the l at the end?

Two impedances in parallel add as:

$$Zp = \left(\frac{1}{Z1} + \frac{1}{Z2}\right)^{-1}$$

so the current through cap will be I = V/Zc?

Sure. What's Zc?

 

[gl0ck]

isnt it capacitor's impedance because the other equation that can be found for current through capacitor is I= C*(dv/dt)

 

[gneill]

isnt it capacitor's impedance because the other equation that can be found for current through capacitor is I= C*(dv/dt)

Yes, what's the formula for a capacitor's impedance?

 

[gl0ck]

so do I need to find it in terms of C or
what is the equation for the current flow through the capacitor since we do not have t..?

 

[gneill]

so do I need to find it in terms of C or
what is the equation for the current flow through the capacitor since we do not have t..?

You wrote the impedance for the inductor as jωL. What's the equivalent expression for the capacitor?

 

[gl0ck]

jwC but I wrote it before ... I think I wrote it somewhere I = V/jwC or it should be I=V/(1/jwC)

 

[gneill]

jwC but I wrote it before ... I think I wrote it somewhere I = V/jwC or it should be I=V/(1/jwC)

No, the expression for the impedance of capacitance C is not jωC. It's 1/(jωC) = -j/(ωC). So your second version for its current is correct.

The potential across it in this problem is E, so the current becomes I_C = E/(1/(jωC)) = jE/(ωC). Edit: Oops, sorry, that should be jEωC.

Now add your two expressions together (add up the real and imaginary components separately).

 

[gl0ck]

It = E/(R+jwL) + jE/(wC)

 

[gneill]

It = E/(R+jwL) + jE/(wC)

Sorry, please see my edit for my previous post. The capacitor current is jEωC.

Now, you went to all the trouble to write the Jacuzzi branch current in the form a + jb before, so why go back to the un-normalized version now? Rewrite the equation above summing the real and imaginary components.

 

[aralbrec]

isnt it when they are in parallel 1/Z_t = 1/Z_c + 1/Z_r+l ?
Zt=(RjwC)/(R+jwC-w^2LCR) ?

The general formula is right but you've made a mistake in your calculation. R+jwL should be in the numerator of Zt.

You can compute the total current both ways. gneill is suggesting adding the two currents together which is another way.

If you find Z_t and look at the magnitude of the current supplied by the source |I| = |V| / |Z_t| you will see a certain value for the capacitor will minimize |I|.

You will see the same result if you find the current through the capacitor and add that to the current through R/L.

====

I just want to post to make sure you are not missing the big picture but if this is confusing you can safely skip it... recall I said an inductor's impedance is complex because it introduces a time delay between the voltage and current waveform. An inductance introduces phase lag which means the current waveform lags behind the voltage waveform. V = I jwL. If you draw that on the complex plane with 'I' horizontal then V will be on the positive imaginary axis so that it is ahead of the current by 90 degrees. This means if the voltage applied to the inductor follows a sinusoid then the current through the inductor will be a sinusoid delayed by 90 degrees of phase shift which corresponds to a delay of t = phase/w seconds. The resistance in series with the inductor changes the phase delay from 90 degrees to something less as you can see from Z = R + jwL. The real part R pushes the impedance jwL off the imaginary axis to the right so the phase delay is less than 90 degrees (ie time delay is less) but it also means the magnitude of the current flowing will be less since |Z| is now √(R² + (wL)²) ohms instead of the smaller (wL) ohms.

On the other hand, a capacitor's current leads the voltage applied to it. V = I 1/(jwC) = -I j/(wC). If you draw that on the complex plane with 'I' horizontal then V will be on the *negative* imaginary axis. This means if a sinusoidal voltage is applied to the capacitor, the current will also be sinusoidal but it will be ahead of the voltage waveform by 90 degrees or ahead by t = phase/w seconds.

Now imagine the same sinusoidal voltage applied to a capacitor and inductor in parallel. The capacitor will draw a current that is ahead of the voltage waveform and the inductor will draw a current that is behind the voltage waveform. The source supplies current and absorbs current from the inductor and it also supplies and absorbs current from the capacitor but because the current drawn for the capacitor and inductor has a certain time relationship (one leads the voltage, the other lags the voltage), the current drawn out of the capacitor can supply the current sunk into the inductor and vice versa. In other words, current can flow back and forth between the capacitor and inductor without coming from the source.

The frequency at which the capacitor and inductor currents can completely supply each other is called resonance. At that frequency, the source supplies no current and sees an infinite impedance (jwL || 1/(jwC)) is infinite.

In your circuit, the source will always have to supply energy to push current through the resistor but a certain value of the capacitance will allow the capacitor to supply all the current demanded by the inductor so that the source does not need to supply it. This will happen when |Zt| is maximized. This is also called power factor correction btw.

 

[gl0ck]

I think I will give up for now .. I looked at it too much that I cannot come up with anything that makes sense ... Thanks a lot for all the effort you made to explain me :)
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL) or

 

[aralbrec]

my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL)

This is right.. keep going when you're ready.

I = E (jwC + 1/(R+jwL)) = E ((jwC)(R+jwL)+1)/(R+jwL)

(I found a common denominator and added)

Multiply that out and see if you can find a C that will minimize |I|. The j*j you see in the numerator above will introduce a negative quantity that will let you make |I| smaller.

 

[gl0ck]

Thanks a lot man but I have to go to sleep.. I hope tomorrow to find the right solution .. For now I give up..
Thanks for the efforts both of you

 

[gneill]

I think I will give up for now .. I looked at it too much that I cannot come up with anything that makes sense ... Thanks a lot for all the effort you made to explain me :)
my last shot is when jwC is equal to jEwC so I(t) = jEwC + E/(R+jwL) or

Good. Separate the total real and imaginary components and factor out the common bits:
$$I = E \left[ \frac{R}{R^2 + (ω L)^2} + j ω \left(C - \frac{L}{R^2 + (ω L)^2}\right)\right]$$
What value of C will minimize the magnitude of the whole?

 

[gl0ck]

Thanks! So when I plugged in the numbers It gets really nasty. First can I use w = 2Pi and to treat jEwC as j220*2*Pi*C? also w*L to be 12.57 because jwL = j12.57 Ohms
so to the equation I = 220 [ R/(R^2+12.57^2) + j2*Pi(C - 0.4/(R^2+12.57^2)]
I= 220[8/(64+158.0049) + j220*6.2830 ( C - 0.4/(64+158.0049)]
I = 7.92+(220jwC - 220*0.0018jw)
I= 7.92 + j1382.26C - j2.4807C
I=7.92+j1379.78C

I hope to do the right calculations

Thanks

 

[gneill]

The impedances of reactive components (inductors, capacitors) change with frequency. Furthermore, they change in opposite directions so that inductor impedance goes up with increasing frequency while capacitor impedance goes down. So no, in general you can't just set ω to a convenient value like $2\pi$; you must use the value specified.

For part (a) you just need to derive the expression for the current, you don't have to plug in any numbers.

For part (b) you need to first determine an expression for C that will minimize the magnitude of the current. Use the given hint.