Is the Solution to This Standardized Test Circuit Problem Correct?

[Firefox123]

Simple circuit problem...

I was lookig at a circuit problem in a standardized test and I think the solution may be incorrect...

Here is the problem...

There is a battery (thus DC), a switch (open) and a resistor in series...followed by a capacitor and a lamp in parallel.

The qusetion asks what happens when the switch is closed and gives the answer as" the lamp will flash on and off due to charging and discharging cycles of the capacitor"...this does not sound correct to me.

The answer I would give is as follows...

When the switch is closed at t=0+ the cap looks like a "short" (since it hasnt built up any resistance via the electric field) and all the current flows to the cap...so the lamp doesn't light.

As time goes on the cap charges and eventually all current to the capacitor stops. So the cap has some voltage across it, which means the lamp (resistor) also has that voltage across it since they are in parallel.

Thus, when the switch is closed, there will be a current flowing through the lamp effectively lighting it.

Now it is true that the cap will try to discharge through the lamp even when the switch is closed...but wouldn't the discharge/charge time for the cap be so fast that there wouldn't be any noticable "flash" for the lamp while the switch is closed?

If anyone can tell me if I am off base here I would appreciate it...

Thanks.

 

[chroot]

There are no charge-discharge cycles. Once the switch is closed, the capacitor will build up an electric field and stay there, stable, until the switch is opened again. If the switch is never opened, nothing ever happens to the capacitor.

- Warren

 

[Firefox123]

There are no charge-discharge cycles. Once the switch is closed, the capacitor will build up an electric field and stay there, stable, until the switch is opened again. If the switch is never opened, nothing ever happens to the capacitor.

- Warren

Thanks for the response...your answer is what I originally thought...

The answer of "charge/discharge" cycles seemed confusing to me, so I tried to imagine a situation where that could happen with the switch closed...which is why I thought even if there was a charge / discharge time it would basically be instantaneous.

I won't be so quick to assume I am wrong and the answer in the back is correct next time...



Russ

 

[Integral]

What if the bulb was had definite turn on voltage, like I believe some of the small neon bulbs. Then the cap would charge until the turn on voltage was reached, the bulb would stay on until the voltage droped below a lower threshold. There is your charge/discharge cycle.

http://www.elecdesign.com/Articles/ArticleID/1032/1032.html

 

[don rigby]

there is a voltage divider cct between the resistor and the lamp/capacitor that will only determine the intensity of light from bulb after the charging of the capacitor to the point where it is effectively fully charged in the circuit, at this time the bulb will stay lit at the percentage of intencity determined by the voltage divider between the bulb and the resistor(capacitor will no longer have an effect). in short switch closes no light, then brighter and brighter until maximum light and equalibrium in intencity. YOU WERE RIGHT!
ps. consider a bulb as resistive unless otherwise specified.
***when the switch is opened however, don,t expect the bulb to go out straight away as it will dim until extinguished.
***guess why! the capacitor is now the the battery (dc source)
***do the math! [with regards to charge discharge times. consider lamp lit until 0 volts (impractical but good exercise)]

 

[Firefox123]

What if the bulb was had definite turn on voltage, like I believe some of the small neon bulbs. Then the cap would charge until the turn on voltage was reached, the bulb would stay on until the voltage droped below a lower threshold. There is your charge/discharge cycle.

http://www.elecdesign.com/Articles/ArticleID/1032/1032.html

I don't think so off the top of my head...but Ill have to look at the problem again. I don't think it was that complicated since its not really intended for someone with a background in EE...

But Ill check and make sure.




Russ

 

[Firefox123]

there is a voltage divider cct between the resistor and the lamp/capacitor that will only determine the intensity of light from bulb after the charging of the capacitor to the point where it is effectively fully charged in the circuit, at this time the bulb will stay lit at the percentage of intencity determined by the voltage divider between the bulb and the resistor(capacitor will no longer have an effect). in short switch closes no light, then brighter and brighter until maximum light and equalibrium in intencity. YOU WERE RIGHT!

I don't understand...are you saying the bulb would flash when the switch is closed or are you saying the bulb would not flash when the switch is closed?

Once the cap has fully charged it doesn't seem like the bulb would flash...

ps. consider a bulb as resistive unless otherwise specified.

Right...that is the general rule I go by.

***when the switch is opened however, don,t expect the bulb to go out straight away as it will dim until extinguished.
***guess why! the capacitor is now the the battery (dc source)
***do the math! [with regards to charge discharge times. consider lamp lit until 0 volts (impractical but good exercise)]

Correct...thats what would happen when the switch opens and the cap begins its discharge through the bulb.



Russ

 

[don rigby]

the bulb will not flash
sorry i did not make that clear
don

 

[The Glom]

Actually, the bulb will produce a very brief, slightly brighter pulse as the switch is opened.

This is because the capacitor has a much lower internal resistance than the battery. So, as soon as the capacitor is allowed to discharge through the bulb, that circuit has only the resistance of the bulb in series with essentially a non-resistive current source, the capacitor.

Try this one.

Place the bulb in series with the capacitor. Draw it so that the bulb is above the capacitor.

Now draw a switch on the right side that connects the top bulb connection to the bottom of the capacitor.

Now, on the left, draw your battery with it's negative lead connected to the bottom of the capacitor. Draw a switch in series between battery positive and the top of the bulb.

Closing the battery side switch will produce a pulse of light that diminishes as the capacitor charges. Notice the intensity and duration are based on the combined internal resistance of the battery and the bulb.

Open that switch and then close the right-side switch. This directly discharges the capacitor through the bulb, producing a somewhat brighter and shorter duration pulse.

 

[don rigby]

"Actually, the bulb will produce a very brief, slightly brighter pulse as the switch is opened."

what you are suggesting contradicts ohms law.

for theoretical purposes:
in order for the bulb to burn brighter on initial switch opening do to internal resistances of the battery, the battery voltage must be dropping due to loading (which i probably agree with to a negligible degree, but only adds confusion to a quite simple concept)
in order to prove theory, we should keep the values understandable (what if we purchased a battery as large as my home town to operate a 40 watt incandescent bulb?).
practical application comes easier in time. let's help this guy crawl!

 

[blue_raver22]

Hmmm, off the top of my head the bulb will turn on upon switching but will die down slower when it is opened again due to the discharge...

that is if memory serves me right