Why is Power Dissipation Divided by Two in AC Circuits?

[amitjakob]

hello all
just studying to that crazy electrical engineering 2 exam coming up friday
encountered something weird about power dissipation.

Given is an AC circuit, source of 10V f=100Hz Z=20 Ohm
Matched to give maximum power is the load impedance which will be 20 Ohm as well.

now my problem:
I have assumed to calculate the current (10V/40 Ohm = 0.25A) and then easily apply I^2*V which gave 1.25W now, in the answer sheet is given solution just half of it : 625mW
now ... why did they divide it by two? how should I know when to do that ? is that a different method I should apply for DC / AC circuits ? does it make any difference that it is called impedance and not ressistance

Thank you all for you time people
Greets, Amit.

 

[amitjakob]

And one more thing later on...

Load of 20 Ohm is actually combined from R||(C+L)
where R=50Ohm C=159.15microF xL=omega*L=10Ohm --> L=pi/20 ?? (omega=200pi)

so, after the dissipated real power over the load has found 625mW , which i still don't know why. we've been asked to find the power over the resistor(50Ohm) which is parallel to the CL combinatioin and is found to be Zero

left me with my mouth open with no direction to start and guess this answer

any ideas ?
sorry for bothering, thanks for reading&answering

 

[ranger]

hello all
just studying to that crazy electrical engineering 2 exam coming up friday
encountered something weird about power dissipation.

Given is an AC circuit, source of 10V f=100Hz Z=20 Ohm
Matched to give maximum power is the load impedance which will be 20 Ohm as well.

now my problem:
I have assumed to calculate the current (10V/40 Ohm = 0.25A) and then easily apply I^2*V which gave 1.25W now, in the answer sheet is given solution just half of it : 625mW
now ... why did they divide it by two? how should I know when to do that ? is that a different method I should apply for DC / AC circuits ? does it make any difference that it is called impedance and not ressistance

Thank you all for you time people
Greets, Amit.

I guess you're finding the power delivered to the load? It seems that 1.25W is is the product of I_RMS and V_RMS. But what if we want to find the power across the 20ohm impedance? Shouldn't it be I_RMS²*20ohm? That gives 625mW.

 

[ranger]

Load of 20 Ohm is actually combined from R||(C+L)
where R=50Ohm C=159.15microF xL=omega*L=10Ohm --> L=pi/20 ?? (omega=200pi)

so, after the dissipated real power over the load has found 625mW , which i still don't know why. we've been asked to find the power over the resistor(50Ohm) which is parallel to the CL combinatioin and is found to be Zero

left me with my mouth open with no direction to start and guess this answer

any ideas ?
sorry for bothering, thanks for reading&answering

Well from the data you've listed, it seems that 100Hz is the resonant frequency for L and C i.e. both the inductor and capacitor have the same reactance (10ohms). When you look at the impedance graph for a capacitor and inductor, what do you notice? Don't they oppose each other? This means...?

 

[amitjakob]

Thanx 4 the answers, ranger !

but still ... few points arnt so clear to me.

1. Irms: is that simply I/sqrt(2) or is it more complicated then that ? when do I use it, in all AC circuits ? in DC
do i use regular current or RMS as well ?

2. I would appreciate if you can put some more light or even link me
to more information about the second point where L&C oppose each
other and there fore no power is dissipated over the parallel resistor.

Thanks for the patience (-:

 

[ranger]

but still ... few points arnt so clear to me.

1. Irms: is that simply I/sqrt(2) or is it more complicated then that ? when do I use it, in all AC circuits ? in DC
do i use regular current or RMS as well ?

Correct. To get the rms current/voltage, just divide by sqrt(2).

Given is an AC circuit, source of 10V f=100Hz Z=20 Ohm

This translates to:
v(t) = 10*sin wt
Given the frequency, one can find omega. The form which its stated above is small signal variation (AC). Here, 10 represents the maximum voltage (peak). It should be obvious why dividing by sqrt(2) gives RMS.
Let me ask you something, of what use is it to use RMS (effective) power as opposed to simply using peak?

2. I would appreciate if you can put some more light or even link me
to more information about the second point where L&C oppose each
other and there fore no power is dissipated over the parallel resistor.

Umm...I can't find a decent link with graph, but how would you express the impedance of both a capacitor and inductor in polar notation (no need to use specific values for Xl and Xc)?

 

[phyind]

well AC means alternating current and DC means direct current. AC changes with time sin wave. thus has frequency while DC is nonfluctuating current and is straight line graph parallel to axis.

Thus when we calculate problems in AC circuits we use rms values called root mean square value as current changes with time (current changes its direction) thus we take average current value to calculate the power same way we use rms values of voltage and thus power=Irms*Vrms/2.

While in DC the current does not change its value and thus we straight way use I value given to us and not rms values.

 

[amitjakob]

Correct. To get the rms current/voltage, just divide by sqrt(2).

This translates to:
v(t) = 10*sin wt
Given the frequency, one can find omega. The form which its stated above is small signal variation (AC). Here, 10 represents the maximum voltage (peak). It should be obvious why dividing by sqrt(2) gives RMS.
Let me ask you something, of what use is it to use RMS (effective) power as opposed to simply using peak?


Umm...I can't find a decent link with graph, but how would you express the impedance of both a capacitor and inductor in polar notation (no need to use specific values for Xl and Xc)?

Hey !
well ... i thinki understand the part where they cancel each other out, but i wonder how and why does it influencing the power on R making it zero
to make it easier i attached the original exam, from it is the first problem

tnx

 

[ranger]

Hey !
well ... i thinki understand the part where they cancel each other out, but i wonder how and why does it influencing the power on R making it zero
to make it easier i attached the original exam, from it is the first problem

tnx

Well if the C and L cancel out each other, then that makes the impedance on the branch which consists L and C...?

 

[amitjakob]

Well if the C and L cancel out each other, then that makes the impedance on the branch which consists L and C...?

0 ... which means all current going through that branch and not through the resistor ?

 

[amitjakob]

Ahhh... i think i can flood this forums now and make its members busy for so long, as long as I am still studying for exam. sun is shining outside, everybody going to swim in the lake ... and I am here.
well at least not alone, huh ?

 

[amitjakob]

i meant these forums and not this forums ... too tired to spell, heheh...

 

[ranger]

0 ... which means all current going through that branch and not through the resistor ?

You can think of it that way. Or you can say since the resistor (R2) is in parallel with 0ohm (a idealized piece of wire), the equivalent impedance is 0ohms. Therefore all the power is dropped across the potentiometer (R1) and the internal source resistance. Since we wish to get maximum transfer of power from source to load, the load should have a resistance equal to that of the source (which you already pointed out). Since this value is 20ohms, it should now be obvious why it seemed like they divided the total power by two.

 

[amitjakob]

Quick problem

From problem no1 in the file attached here.

1e) Which capacitance to make the apparent power real.

My direction was to try and make the Total impedance real, which would make
the Current real (voltage is real).
but I am kinda stuck, not able to proof in this way, I am pretty sure its an 'kinda-easy-if-you-get-the-hang-of-it' question ... but, i guess i didnt

any ideas ?

 

[ranger]

From problem no1 in the file attached here.

1e) Which capacitance to make the apparent power real.

My direction was to try and make the Total impedance real, which would make
the Current real (voltage is real).
but I am kinda stuck, not able to proof in this way, I am pretty sure its an 'kinda-easy-if-you-get-the-hang-of-it' question ... but, i guess i didnt

any ideas ?

Interesting problem. Are you aware of the following:
$$S = P + jQ$$
where S is apparent power, P is real power and Q is reactive power.

So you are trying to make S = P right? You can read this for more help on the issue:
http://www.ibiblio.org/obp/electricCircuits/AC/AC_11.html

 

[amitjakob]

Thanks for the link, it looks interesting, just started to read
one thing bothered me there
They relate to amplitude of V or I as RMS, just as it is and without dividing
it by sqrt(2)for example, at the explanation coming directly after the RL series circuit
they say its a source of 120V RMS ... or does that mean it actually has the following formula v(t)=120*sqrt(2)*cos(wt+a)
or am i just competly mixed ... hopefully reading on will clear some stuff.

 

[ranger]

Thanks for the link, it looks interesting, just started to read
one thing bothered me there
They relate to amplitude of V or I as RMS, just as it is and without dividing
it by sqrt(2)for example, at the explanation coming directly after the RL series circuit
they say its a source of 120V RMS ... or does that mean it actually has the following formula v(t)=120*sqrt(2)*cos(wt+a)
or am i just competly mixed ... hopefully reading on will clear some stuff.

The highlighted section implies that you are indicating a time varying signal - not RMS! That 120V volt in that expression is the peak.

When we work with AC circuits, and we need to work with ohms law, power law, voltage divider, etc, we must use complex notation. A majority of these require use to divide and multiply quantities. You should know that polar notation is best suited for this. And of course the magnitude in polar notation is always RMS.

EDIT: sorry. I missed when you multiplied 120V*sqrt(2). In that case, its in peak. And it fits in properly to express small signal variations. So you're correct.

 

[amitjakob]

Very intereseting article, so much that i would never get from
my sleepy self-bored-to-death professor.
Anyways, too much already over 10 hours today
so i'll retire for now, see you tomrrow (for sure! hehehe...)

Night