# Homework Help: Problem with half power frequencies ?

1. Oct 23, 2011

### the_kool_guy

there is a parallel R - C circuit connected with a alternating current source of frequency 'w'.
on plotting the function of |z| vs 'w' we get max at w=0; and max/2 at 'w'= 1/RC.
while calculating power absorbed by resistor,
we get Pmax = 1/2I^2R( >>>? how)..
and at w = 1/RC power by resistor is Pmax/2.
...
,,,
my question is how can Pmax be 1/2I^2R as power is max at w = 0 and at that its a DC where
power max must be I^2R...
can some1 enlighten me in this topic pls...

also at w = 1/RC,power is 1/2 of above defined power which is easily calculated.. so it is termed as half power freq..
..?>>
thanks

2. Oct 23, 2011

### Staff: Mentor

At ω = 1/RC, the |z| should be max/√2.

In order to show that the power dissipated by the resistor is half the maximum I2R when ω = ωo = 1/RC, you can derive an expression for the current in the resistor (the RC circuit is behaving as a current divider), and then square that current and multiply by the resistance. What's the magnitude of the power?

Last edited: Oct 23, 2011
3. Oct 23, 2011

### the_kool_guy

it will be half of I2R but aren't we talking about average power dissipated or rather its instantaneous power that must be taken to calculate half power frequencies?

4. Oct 23, 2011

### the_kool_guy

here's the snap of text m consulting...... its calculated for average power.......
even for w = 0, average power is said to be 1/2I*I*R

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5. Oct 23, 2011

### Staff: Mentor

If your current is specified as an RMS value then the power calculated will be RMS, too. If you like doing the math you can take I as a peak value and then calculate the instantaneous current through the resistance, then find the instantaneous power, and finally use an integral to calculate the RMS value. Seems like a lot of work to me

6. Oct 23, 2011

### the_kool_guy

my question is that...
AVERAGE power when w = 0 should be I*I*R.
and when w = 1/RC Average power comes(by resistor) to 1/4*I*I*R...
this thing is confusing me as the text declares AVERAGE power with w = 0 as 1/2*I*I*R

7. Oct 23, 2011

### Staff: Mentor

Ah. Well, if V is specified as RMS, then when ω = 0 the continuous voltage applied will be the PEAK value, or √2*V. Square that and you get 2*V2. The "1/2" in (1/2)V2/R takes care of cancelling with the "2" in the square of the voltage.

8. Oct 23, 2011

### the_kool_guy

In the text there is no mention of RMS .
there is just written for I.i am writing the complete statement of text..

GIven the current I,then |V| is max when |z| is max or w = 0 where |Z| = R so avg power absorbed by resistor is Po=1/2|V|^2/R....

however when w=1/RC then |Z| = R/sqrt(2)...P1=1/4*|I|*|I|*R=Po/2..
....

pls explain

9. Oct 23, 2011

### Staff: Mentor

Average power for an AC circuit is given by $P = V\overline{I}$, where V is the RMS voltage and $\overline{I}$ is the complex conjugate of the RMS current.

Suppose that the peak value of the current is given as I. Then its RMS value is I/√2. Now let's look at the impedance and voltage across the RC circuit.

The impedance is Z = R || Zc, where Zc is the impedance of the capacitor. In this case we're interested in the impedance when ω = 1/(RC), which makes Zc = -jR. The impedance of the circuit is thus

$Z = \frac{R}{2}(1 - j)$

The voltage across the RC pair will be:

$V_{RMS} = I_{RMS}Z = I_{RMS}\frac{R}{2}(1 - j)$

The current through the resistor in particular will then be:

$I_R = V_{RMS}/R = \frac{I_{RMS}}{2}(1 - j)$

The power in the resistor is then

$P_R = V_{RMS} \overline{I_R} = \left(I_{RMS}\frac{R}{2}(1 - j)\right) \left( \frac{I_{RMS}}{2} (1 + j) \right)$

$P_R = \frac{1}{2} (I_{RMS})^2 R$

But IRMS is I/√2, so

$P_R = \frac{1}{4} I^2 R$

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