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# Resistance wattage in series

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Hi
I analyze a timer circuit using transformerless power supply, that is no transformer, but a 330 nF in series with two 240 ohm 1/2 watt resistors.
i noticed that the designer always use two-series 240 ohm 1/2 watt resistors. he do so three times in different places in the circuit.
Also he connect the emitter of a small signal transistor (BC547B) to ground through three-series 1Kohm 1/2 watt.
Why the designer do this?
what the equivalant wattage of two 1/2 watt resistors connected in:
(a) series
(b) parallel
thank u very much

A = 1/2W
B = 1W

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Are u sure Ante, or, can I say that three 1K ohm 1/2 watt resistors in series are equivalent to one resistor of 3k ohm 1/2 watt
if yes please tell me to be sure
thanks

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Resistor wattage is always added together. No matters is it series or parallel connection.

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Oh what would i do now
this i different saying
now i need a third person to put me in the correct road, hope to be audioguru.

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What did Ante say about resistors in series? Each resistor dissipates 0.5W.

If two series resistors have the same value then each one dissipates half as much as if you used a single resistor that has twice the resistance of each one, because they share the power. Just like they are in parallel where they share the power to equal a resistor with half the resistance of each one.

Three 1k, 0.5W resistors in series are equal to one 3k, 1.5W resistor.
Three 1k, 0.5W resistors in parallel are equal to one 333.3 ohm, 1.5W resistor.

Sometimes resistors are used in series to increase the max voltage that can be safely applied. Weren't you talking about a 240VAC mains, that has a peak voltage of 340V?
A little resistor might arc with such a high voltage across it. Check the rating on its datasheet. ;D

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Hi
The circuit shown below is the power supply section of a stairs light timer. I draw it from a real timer circuit. I have some questions.
(1) The 18K-1/8w resistor. It waste (9V/18K=) 0.5mA current to ground all the time. What the purpose of this resistor?
I check its connection many time and I'm sure it is connected in the place shown in the figure.
(2) Why the designer uses two series 240 ohm - 1/2 watt resistors? Why not one 470 ohm 1/2 watt? What the difference? or what is his other choice?
(3) The green led limiting the current to 20mA, i guess it i am not sure!

thanx

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I have some questions.
(1) The 18K-1/8w resistor. It waste (9V/18K=) 0.5mA current to ground all the time. What the purpose of this resistor?

Maybe it is needed to discharge the 10uF cap quickly when the power is cutoff.

(2) Why the designer uses two series 240 ohm - 1/2 watt resistors? Why not one 470 ohm 1/2 watt? What the difference? or what is his other choice?

Two 240 ohm - 1/2W resistors in series is equal to a 480 ohm - 1W resistor. Maybe the designer had lots of 240 ohm - 1/2W resistors. Maybe he needs 480 ohms but only 470 ohms is available.

(3) The green led limiting the current to 20mA, i guess it i am not sure!

No, LEDs don't limit current. The 240 ohm amd 10k resistors and the reactance of the 330nF cap limit the current.

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hi audioguru
the answers of part 2 and 3 are ok
the answer of part 1 is not satified me, i don't think it for dischage the 10u cap

what if i remove it from the circuit what would u expect?

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i guess what bloki said is not correct.
wattage of the resistors is added only when they are connected in parallels and not in series.
whats the view of others?

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the answer of part 1 is not satified me, i don't think it for dischage the 10u cap

what if i remove it from the circuit what would u expect?

The 18k capacitor discharges the 10uF capacitor in 180ms, reducing damage to the LED when its voltage is reversed when the power is turned off. The absolute max reverse voltage for most LEDs is only 5V, and this circuit has 9V.

I never operate parts beyond their max ratings so I don't know how soon the LED will fail due to extended time of reversed over-voltage if you remove the resistor.
The LED should have a diode connected in reverse in parallel with the LED to protect it. Then removing the resistor won't make much difference.

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i guess what bloki said is not correct.
wattage of the resistors is added only when they are connected in parallels and not in series.
whats the view of others?

Of course the wattage of the resistors is doubled when they are in series. Two 240 ohm resistors are 480 ohms. Put 22V across both in series and the total dissipation is 1W which is divided between the resistors because their value is the same. Therefore two 240 ohm half-watt resistors in series is the same as a 480 ohm 1W resistor.

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The 18k capacitor discharges the 10uF capacitor in 180ms, reducing damage to the LED when its voltage is reversed when the power is turned off. The absolute max reverse voltage for most LEDs is only 5V, and this circuit has 9V.

I never operate parts beyond their max ratings so I don't know how soon the LED will fail due to extended time of reversed over-voltage if you remove the resistor.
The LED should have a diode connected in reverse in parallel with the LED to protect it. Then removing the resistor won't make much difference.

this is excellent answer thank you audioguru

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hi audioguru!
say we have 2 resistors with wattage P1 and P2 and connected in series.
the current through them is same i=V/(R1+R2)
net resistance=R1+R2
in such a case it can be written as
V*V/P=V*V/P1+V*V/P2.
thus 1/P=1/P1+1/P2
thus reciprocals added give reciprocal of net power.what do u say?

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Hi Prateek,
When resistors in series have the same value, their power rating is simply added. Why use recipricals?

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hi audioguru!
what will be the case according to u when the resistors are in parallels?
then i guess current through them being unequal and voltages being equal we would have P=P1*P2/(P1+P2) as the net power rating where P1 and P2 are the individual power ratings.
am i right?

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