Q&A - PSLE Science

[tianzhu]

chrisu:

My daughter gave the same answer - (2) too.


But I think if Both Switches A and B are closed, wouldn't the Switch A caused a short circuit to the bulb besides Switch B? So only 1 bulb will be lighted up.

Hi chrisu

Good Morning.

In arrangement 2 , you have a single battery connected to two parallel circuits.

Visualise as one circuit with one bulb as compared to another circuit with two identical bulbs.

The sum of currents entering any junction in a circuit is equal to the sum of currents leaving the same junction.

Hence two bulbs will be turned on.

Interesting, you’ve brought up short circuit which I think is not in the PSLE science syllabus.

If one of the paths in a circuit is without a light bulb, the path has very little resistance and virtually all the current take this path. Thus, a short circuit occurs.

Please go to page 29 and refer to Q43 of this testpaper.
http://www.wendykoh.com/08/primary6-nanyangca1-science.pdf

A short circuit occurs when switch D is closed. Notice that the circuit consists of a conducting wire without light bulbs.

You may refer to this link to learn more about short circuit.
http://www.furryelephant.com/content/electricity/complete-circuits/short-circuits/

Hope this helps.

Best wishes.

[chrisu]

Hi Tianzhu,


Good morning to you too, you are pretty early. Thank you for your reply and I do agreed that short circuit may not be a good term to use for Primary school since it's not in syllabus.

In order for a bulb to light up, there must be a battery source (potential difference though again not in primary syllabus) connected to it correctly. That is the +ve of the battery connected to one point of the bulb and the -ve of the same battery to the other point of the same bulb. Diagram (c) illustrate this point. (sorry for the poor drawing). Technically, there is a battery source (potential difference) connected to the bulb correctly.

http://img.photobucket.com/albums/v160/guppiesonli/Misc/P6-Sc2.jpg\">

Diagram (b) light bulb will not light up as the two points of the bulb is connected to the same point (-ve) of the battery. Technically there is no potential difference between the two points of the bulb or the bulb is not connected correctly to the battery.

Diagram (a) shows the answer (2) redrawn to illustrate that only 1 bulb will light up if both switches A and B are closed. For easier understanding, I'm renaming the bulb besides the switch B as Bulb B and the other bulb as Bulb A. The arrow shows the same wiring in the following diagram but drawn differently. In the last diagram of (a), it shows clearly why the Bulb B will not light up because the battery source is not connected to it correctly or no potential difference across it.

As for the link to the Question 43 paper, the model answer is Potential energy ---> Electrical energy when all swithes are closed do agreed with my point. When all swithes are closed, the other appliances will not work thus no light, sound and heat energy converted and only electrical energy is converted as the wire is connected between the +ve and -ve of the battery source. There will be current flowing thru it (wire) as it's a complete circuit. Sad to say it didn't explain why the other 3 appliances has no energy conversion (will not work) since short circuit is not in syllabus.

It will be better if the teacher will to set the question in such manner that the switch is closed individually only since it will still explain the various energy conversion for primary school.

Hope it helps.

tianzhu:

chrisu:

My daughter gave the same answer - (2) too.

But I think if Both Switches A and B are closed, wouldn't the Switch A caused a short circuit to the bulb besides Switch B? So only 1 bulb will be lighted up.

Hi chrisu

Good Morning.

In arrangement 2 , you have a single battery connected to two parallel circuits.

Visualise as one circuit with one bulb as compared to another circuit with two identical bulbs.

The sum of currents entering any junction in a circuit is equal to the sum of currents leaving the same junction.

Hence two bulbs will be turned on.

Interesting, you’ve brought up short circuit which I think is not in the PSLE science syllabus.

If one of the paths in a circuit is without a light bulb, the path has very little resistance and virtually all the current take this path. Thus, a short circuit occurs.

Please go to page 29 and refer to Q43 of this testpaper.
http://www.wendykoh.com/08/primary6-nanyangca1-science.pdf

A short circuit occurs when switch D is closed. Notice that the circuit consists of a conducting wire without light bulbs.

You may refer to this link to learn more about short circuit.
http://www.furryelephant.com/content/electricity/complete-circuits/short-circuits/

Hope this helps.

Best wishes.

[tianzhu]

chrisu:

Hi Tianzhu,

In order for a bulb to light up, there must be a battery source (potential difference though again not in primary syllabus) connected to it correctly. That is the +ve of the battery connected to one point of the bulb and the -ve of the same battery to the other point of the same bulb. Diagram (c) illustrate this point. (sorry for the poor drawing). Technically, there is a battery source (potential difference) connected to the bulb correctly.

Hi chrisu

Good Morning.

I am thinking about children doing experiment using miniature light bulbs with miniature plastic bulb holders.

The configuration is as shown in the image. I am not too sure if it works, as I don’t have a set of kit to carry out the experiment at home.

http://farm7.static.flickr.com/6121/6025190684_1f4a3c9165_z.jpg

I see the schematic diagram of option (2) as two parallel loops sharing a single source.

Loop A has one bulb A and loop B consists of bulb B as well as bulb A.
The sum of currents entering any junction in a circuit is equal to the sum of currents leaving the same junction. In this case, more current is passing through bulb A, hence it’s brighter than bulb B.

Best wishes

[chrisu]

Good morning Tianzhu.


Thanks for the pictorial diagram. I've redrawn it and it's still the same as yours just that the point of the brown wire is now at the same positive terminal of the battery. The bulb B will not light up is both its wire is connected to the same positive point of battery.

http://img.photobucket.com/albums/v160/guppiesonli/Misc/Bulb2.jpg\">

tianzhu:

chrisu:

Hi Tianzhu,
In order for a bulb to light up, there must be a battery source (potential difference though again not in primary syllabus) connected to it correctly. That is the +ve of the battery connected to one point of the bulb and the -ve of the same battery to the other point of the same bulb. Diagram (c) illustrate this point. (sorry for the poor drawing). Technically, there is a battery source (potential difference) connected to the bulb correctly.

Hi chrisu

Good Morning.

I am thinking about children doing experiment using miniature light bulbs with miniature plastic bulb holders.

The configuration is as shown in the image. I am not too sure if it works, as I don’t have a set of kit to carry out the experiment at home.

http://farm7.static.flickr.com/6121/6025190684_1f4a3c9165_z.jpg

I see the schematic diagram of option (2) as two parallel loops sharing a single source.

Loop A has one bulb A and loop B consists of bulb B as well as bulb A.
The sum of currents entering any junction in a circuit is equal to the sum of currents leaving the same junction. In this case, more current is passing through bulb A, hence it’s brighter than bulb B.

Best wishes

[chrisu]

I've purchased these bulbs from Sim Lim just for experiment for my daughter to visualise better.


The first one below shows both bulbs light up equally though picture may not show equal brightness due to camera. This is the same before Switch A is closed (Just Swithc B closed);
http://img.photobucket.com/albums/v160/guppiesonli/Misc/BulbA.jpg\">

Now with both Switch A and B closed, only 1 bulb is lighted up (very bright) and the other buld is not lighted at all;
http://img.photobucket.com/albums/v160/guppiesonli/Misc/BulbB.jpg\">

Hope it's clear. Thanks.

[tianzhu]

Hi Chrisu


Thank you for carrying out the experiment. This means none of the possible answers given in the test paper is correct.

http://img.photobucket.com/albums/v160/guppiesonli/Misc/P6-Sc.jpg

I apologise for leading you the wrong way.

You may wish to check with your kid’s teacher for the right answer.

Best wishes

[chrisu]

It's alright Tianzhu. I will get my daughter to check with the teacher.


I understand why most ppl choose (2) as answer; It is true that total current flow to that junction equal to the sum of both flowing out. IT = Ir + IR;
http://img.photobucket.com/albums/v160/guppiesonli/Misc/Calculating.jpg\">

Let us not forget that when Switch A is closed, the wire resistance is probably very low (can assume to be 0.0001ohm). You may read here for http://www.furryelephant.com/content/electricity/resistance-ohms-law/.

So the total resistance of the wire (when switch A closed) and bulb (assume to be 10ohm) in parallel = 0.0000099 ohm.
Total resistance in circuit = 10 + 0.0000099 = 10.0000099 ohm.
Therefore total current IT = 3/10.0000099 = 0.2999997 A

Voltage across the first bulb connected nearest to battery = IT x 10 ohm
= 0.2999997 x 10 = 2.999997V

As you can see, most of the 3V battery is dropped across this bulb (2.999997V) and very very little voltage across the other bulb - thus the voltage is not high enough to light up that bulb.

tianzhu:

Hi Chrisu

Thank you for carrying out the experiment. This means none of the possible answers given in the test paper is correct.

http://img.photobucket.com/albums/v160/guppiesonli/Misc/P6-Sc.jpg

I apologise for leading you the wrong way.

You may wish to check with your kid’s teacher for the right answer.

Best wishes

[atutor2001]

Hi Chrisu


I didn't pay attention to the question till you posted your illustrations and experiments.

You have very good electrical knowledge. Many will have great difficulty visualising the short-circuit caused by switch A on bulb B - few will know how to flip switch A down like you did in your diagram. I believe the pr sch teacher who set the question didn't realise that too. There are many such mistakes in Pr Sc questions. Yes, you are absolutely correct, none of the circuits fit the results given.

Your child is very lucky to have a parent like you.

Cheers.

ps

Just saw your latest circuit diagram. You are guaranteed electrical engineering trained.

[pixiedust]

Oh dear! I just revised electricity but I am seeing even more stars now! Chrisu, agree your child is really lucky! Appreciate to know (1) where in sim lim to get the experiment kits? (2) please update us when the teacher response regarding the question, thanks very much.


To all : about short circuit - is it not in psle syllabus? I came across one oe question in p5 process skills assessment book and also one paragraph explanation in mc psle science guide.

[chrisu]

Thanks atutor2001. I've studied Electronics in poly many years back (70s) and have forgotten most of it.


I remember I've taught my eldest son in sec.4 last year on bulb brightness in series and parallel circuit (one of those favourite topic in school) and some other bulb questions, and now my daughter P6 also learning the same thing in the worksheet and test.

Primary school text book and what teacher taught are always very brief but the exam and test or worksheet are very difficult.

atutor2001:

Hi Chrisu

I didn't pay attention to the question till you posted your illustrations and experiments.

You have very good electrical knowledge. Many will have great difficulty visualising the short-circuit caused by switch A on bulb B - few will know how to flip switch A down like you did in your diagram. I believe the pr sch teacher who set the question didn't realise that too. There are many such mistakes in Pr Sc questions. Yes, you are absolutely correct, none of the circuits fit the results given.

Your child is very lucky to have a parent like you.

Cheers.

ps

Just saw your latest circuit diagram. You are guaranteed electrical engineering trained.