capacitor in series with a light bulb

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Tracid

Member level 3 capacitor light bulb

Hi!
I just would have a simple question:

i connect a capacitor and a light bulb all in series with DC power supply and GND.

I have learned that capacitor in series with the DC voltage blocks the voltage.

Then why should the light bulb light up for a moment when power is switched ON?

mostafa0020

Member level 1 capasitor series with bulb

The steady state value for current in a DC Circuit composed of a capacitor series with resistor " bulb " is zero, but, during the transiet period, the current has an exponential decay from a peak value " =E/R " and decays to reach zero in some multiples of a certain time constant " τ = RC " so , when switching ON, the bulb will light form some seconds " or minutes depending on capacitor size and bulb resistance ".
for detailed mathematical derivation, refer to any electric circuits handbook like Dorf.

Ashish.chip

Newbie level 6 light bulb in series with a capacitor

hey really this is a confusing question for beginners...i can explain this in many ways but here i want to expalin this in term of fourier tarnsform....u know a DC source have fourier taransform(Frequency Spectrum) like a impuls...

Fourier(DC=1)=2Πδ(ω)
where δ(ω) is called impulse function
ω=frequency
δ(ω)=1 for ω=0;
=0 otherwise
So just for a time instant it shows a voltage..which ons the bulb..........
if u can't satisfy with this ans tel me......i'ld like to explain u in other way...

u know when u connect a capacitor and a resistor(bulb) with a DC source....your syatem will have teo kind of responses
(1)Transient response..............
Voltage across capacitor Vc(t)=Vo[1-exp(-t/RC)]
Voltage across capacitor Vc(t)=Vo
Where V0=Peak voltage in the circuit

So now what do u feel.....u see the bulb on in Transient response of the circuit...so increase time constant(RC) if u want to see transient response more closely...........

jasmin_123

Full Member level 3 light bulb transient response

Tracid said:
Then why should the light bulb light up for a moment when power is switched ON?
To charge the capacitor. Tracid said:
I have learned that capacitor in series with the DC voltage blocks the voltage.
Do not learn; THINK!

Tracid

Member level 3 simple rc circuit with a light bulb

Sorry Ashish.chip your explanation is very difficult for me.
I would need a simple answer/explanation.
When theres a circuit with both AC and DC components in it then the cap will block the DC and pass the AC,no? and this is with simple DC circuit,no?

this is what i dont understand.then if it charges the cap then discharges,this sequence should repeat and thus make some noticeable change in bulb illumination periodically,no? sorry i am really a beginner.

szekit

Member level 5 capacitor and a bulb in series

i = C * dV/dT

sudden change in voltage across the cap will produce current to light the bulb for a moment.

Tracid

Member level 3 capacitor and lightbulb in series

OK,and this is a single action. but why isnt it repeating continuously? why does it(the bulb) stays OFF for eternity?

student153

Member level 1 light bulb by capacitor

when current flows initial through the capacitor it acts as a short circuit initally(momentarilly) but when it chares up then it remains open so initally it conducts but finally it blocks for further elaboration see "david irvin"

joyu

Newbie level 5 capacitor for dc lightbulb

Remember in physics, a voltage ACROSS a cap can not change instantaniously. So when you apply your voltage the DROP across it cannot change

So when you turn on the source both terminals of the capacitor are at the high voltage.

Then as your capacitor charges your node on the bulb side goes to ground.

bluesmaster

Member level 3 lightbuld series

just like pop killer technology in a audio driver

Kral Tracid,
If we consider the bulb to be a constant resistance (which it is not), then the current thru a series RC circuit IC circuit is givern by
I = Imax[exp-(t/RC)].
Imax is the initial current, which is equal to V/R.
At t= 0, the exponential term = 1.
At t=infinity, the exponential term = 0.
At other times, the current decays exponentially.
The real world situation is not quite so simple, because, as the bulb heats up, its resistance increase dramatically.
Regards,
Kral

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