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Author Topic: Power Consumption Calculation  (Read 3005 times)
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thetrueman
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« on: December 26, 2007, 06:02:25 06:02 »

Good Day,

Here is a simple RC series circuit. I want to calculate its actual Power Consumption but I’m little bit confused or I’m forgetting some formula to make calculation.

The current which I measured with meter is 20mA. So the Power is 4.4W but that is for purely resistive circuit.

In series circuit the current is same in the circuit. So I connected 1 ohm resistor in series and measured voltage across the resistor which were 60mV. According to this the current flow in the resistor should be 60/1=60mA but with meter it is 20mA.

I’m sure its reason is series capacitor. So I’m confused how to measure the real power consumption in the resistor. Is there an easy way to calculate power? Thanks.

Shahzad
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neelandan
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« Reply #1 on: December 26, 2007, 12:06:49 12:06 »

Measure the voltage across the resistor. The power is V**2/R.

or measure the current through the resistor. The power is R * I**2

This is true, even if the resistor is in a series or parallel or compound circuit. The problem in power measurement arises only when both terminals of the resistor are not accessible, or if the voltage across is not pure dc or sinewave or an easily measureable mixture of both.
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kahramane
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« Reply #2 on: December 26, 2007, 06:38:54 18:38 »

you cannot measure current in series RC circuit which supplied pure DC, current is always zero. Bacause capacitor has infinite resistance(capacitance) on direct current.

When a sinusoidal signal applied to capacitor it has a capacitance depend to frequency of sinusoidal signal.

The capacitance of a capacitor is Xc = 1 / 2 * PI * f * C

Xc is ohm
PI = 3.14
f = frequency  as Hertz
C = capacity as Farad

if there is a RC serial circuit the equal impedance of circuit(Z) calculated as :

Z = SQRT ( R*R + Xc * Xc)

Now you can calculate current from I = V / Z

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DTiziano
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« Reply #3 on: January 31, 2008, 08:44:57 20:44 »

you cannot measure current in series RC circuit which supplied pure DC, current is always zero. Bacause capacitor has infinite resistance(capacitance) on direct current.

When a sinusoidal signal applied to capacitor it has a capacitance depend to frequency of sinusoidal signal.

The capacitance of a capacitor is Xc = 1 / 2 * PI * f * C

Xc is ohm
PI = 3.14
f = frequency  as Hertz
C = capacity as Farad

if there is a RC serial circuit the equal impedance of circuit(Z) calculated as :

Z = SQRT ( R*R + Xc * Xc)

Now you can calculate current from I = V / Z




"The capacitance of a capacitor is Xc = 1 / (2 * PI * f * C)"

Wich is the wave shape that drive your RC circuit ?
The theory is wright, so to only solution is that your data are wrong.
Take care that low cost meter are calibrated for pure sine wave and the reading get wrong
with different shape and/or at high frequency. Even the expensive true RSM one are not precise at crest factor higher then 4 or 6.



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thetrueman
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« Reply #4 on: February 03, 2008, 12:16:06 12:16 »

In my case power supply is AC220V 50Hz sinwave. So please make calculation as shown in previously posted circuit.

Acctually I saw a LED night lamp as circuit above with a series capacitor and resistor and bridge diode and there was mentioned 120V 60Hz 0.3W 25mA on it. how they calculate power 0.3W?
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V.I.P
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« Reply #5 on: February 03, 2008, 01:01:06 13:01 »

In the case of your night light circuit the bridge rectifier is converting the ac to dc so you can calculate the power dissipated in the circuit by measuring the voltage and current after the rectifier and using P= V*I . If the resistor is before the bridge rectifier the you will have to use ac meters and measure the V and I across the resistor and use P=V*I. Capacitors (ideal) are lossless components and don't dissipate any power (they don't get hot) but they do have impedance and there is a voltage drop across them  so reducing the voltage to the led. There will be a small power loss in the bridge rectifier as well. You could calculate the power from the forward voltage drop multiplied by the dc current. Hope this is clear.
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thetrueman
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« Reply #6 on: February 06, 2008, 10:20:43 10:20 »

Here is simple circuit of 220VAC operated LED. So please can anybody explain the actual power consumption. And how to measure cos phi for capacitor and what to do with Resistor.

Shahzad
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zuisti
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« Reply #7 on: February 06, 2008, 05:48:46 17:48 »

Here is simple circuit of 220VAC operated LED. So please can anybody explain the actual power consumption. And how to measure cos phi for capacitor and what to do with Resistor.

Well, a little arithmetic (from the school) ...
We are using the equates from the posts above.

The .27 uF (2.7e-7 Farad) capacitor has an impedance at 50 Hz:
  Xc = 1/(2 * PI * f * C) = 1/(2 * PI * 50 * 2.7e-7) = 11,79 kOhm = 1.179e+4 Ohm
We calculate with the (rot) LED forward voltage  = ~1.6v, the two Graetz bridge diodes = 2 * .7 V, so
 the sum is 1.6 + (2 * .7) = 3 V. At 20 mA the circuit is assumed as a resistor
 with value R = U/I = 3/.02 = 150 Ohm
The whole (sum) R in the circuit is 1k + 150 Ohm = 1.15e+3 Ohm = 1.15 kOhm.

Now we can calculate the complex impedance:
 Z = SQRT ( R * R + Xc * Xc) = sqrt(1.15e+3 * 1.15e+3 + 1.179e+4 * 1.179e+4) = sqrt (1.403e+8) = 1.185e+4 Ohm
 The effective current,  I = U/R = 220/ 1.185e+4 =  1.857e-2 = 18.57 mA.

The effective power is dissipated on the resistive devices only (sum 1.15 k, see above) beacuse the C is really lossless.
 So P = I * I * R = 1.857e-2 * 1.857e-2 * 1.15e+3 = 0.397 W = 397 mW

About the 1 k resistor in the circuit:
It is a must beacuse it decreases the worst case peak current. If the circuit is switched on at the moment of the peak input voltage (in this case it is 310 v !!), and the capacitor is empty (working as a shortcut), the current is limited with the resistive devices only). Calculate with R = 1.15 k, the peak current:
 Ip = U/R = 310 /1.15e+3 = 0.2695 A = 269.5 mA !!

About the cos PHI ... you can it calculate if needed :-)

An other thing: use a 470k.. 1MOhm resistor parallel with your circuit to discharge the capacitor after the switching off!

Hope this helps
zuisti
 
« Last Edit: February 06, 2008, 06:07:15 18:07 by zuisti » Logged
thetrueman
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« Reply #8 on: February 07, 2008, 06:45:09 06:45 »

Dear Zuisti,

Thanks a lot to solve my problem. Actually last weak I saw another such simple device which was mentioned 0.5W at 220V. So it increases my interest to find the magic of calculation. Now I've found it very clearly. The capacitor is really a  usefull component.

Best Regards,

Shahzad  Cheesy
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armandiaz
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« Reply #9 on: April 18, 2008, 07:10:24 19:10 »


Just to test what really happens I suggest you to download Switchercad from Linear:

http://www.linear.com/designtools/software/#Spice

It is a free sofware able to easy simulate and show the main value of an electric circuit.
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