Q&A - PSLE Science

[Chenonceau]

My son thinks it is (4) too. Anyone else?

[Sun_2010]

pixiedust:

Thanks, Chenonceau & Sun_2010. Appreciate that.


How about this one :
http://i1197.photobucket.com/albums/aa424/pixie_dust8/sci2.jpg\">

The answer is (2) but I can't reconcile it! I thought the answer is (4) ?

Book answer is correct.
Only B2 and B3 can work . Look carefully, B4 is having a wire across it and when the switch is closed B1 will have it too.

[pixiedust]

Sorry, don’t quite get the ‘wire across it’ ?

[Chenonceau]

We also very :?

[cimman]

http://i56.tinypic.com/246nuj7.jpg\">

here's a diagram to explain how the current flows.

The key point to note here, is that current always prefers a path of least resistance. Given a choice, a path of little resistance and a path of large resistance (the bell constitute a path of large resistance), most of the current will flow along the path of least resistance. Whatever miniscule current that flows through the Bell B4 and B1 is insufficient to activate the bell, thus practically no current flows here.

If 2 electrical paths provide similar resistance, then current will flow through both paths.

In the above case, it is important to know that a straight wire (LM and GH), with no device on it's path, will present practically zero resistance to the electric current. Thus, almost all current will flow on this path (the blue path) and ignore the red path.

The concept of resistance is a very important concept to understand in electrical circuits. A device, either a light bulb, a bell, etc.. presents a large resistance to the current. A path which has no such devices, ie. just a straight wire, will present practically zero resistance to the electrical current. Given a choice, current will always flow along the path of zero resistance and ignore the path of high resistance. This is where the term, short circuit, comes in. The straight wire will divert all current to it's path.

[Chenonceau]

Cimman!!! I looooooooooooooooooooooooove you!! Thank you so much for taking the trouble to do up such a detailed and clear explanation. My son is gonna be thrilled when he sees this.

[atutor2001]

cimman:

.... A device, either a light bulb, a bell, a coin, a nail, etc.. presents a large resistance to the current. A path which has no such devices, ie. just a straight wire, will present practically zero resistance to the electrical current. Given a choice, current will always flow along the path of zero resistance and ignore the path of high resistance. This is where the term, short circuit, comes in. The straight wire will divert all current to it's path.

Hi cimman, this is the best diagram I have seen for explanation for on short-circuit. Maybe school textbooks should have it. Thank you.


Incidentally, I think coin and nail have very low resistance and will cause short-circuit.

[Fat Mama]

Hi Cimman,


Just to clarify as I am not v gd at circuits. DD’s teacher said electricity flows from -ve to +ve pole. However in ur diag, the current flows from +ve to -ve. So who is correct?

Is it right to say B1, B2 n B4 are connected in series? B2 n B3 are connected in parallel. So resistence of B1, B2 n B4 are greater that B2 n B3 so the current flows to B2 n B3?

<In the above case, it is important to know that a straight wire (LM and GH), with no device on it’s path, will present practically zero resistance to the electric current. Thus, almost all current will flow on this path (the blue path) and ignore the red path.>

U said LM and GH with no device on it’s path, but isn’t B3 connected to LM and GH?

Tks.

[Chenonceau]

atutor2001:

Incidentally, I think coin and nail have very low resistance and will cause short-circuit.

Can I bother you to explain a bit more?

[atutor2001]

Chenonceau:

atutor2001:

Incidentally, I think coin and nail have very low resistance and will cause short-circuit.

Can I bother you to explain a bit more?

A coin or a nail will allow current to flow through as easily as a wire. Therefore, if we connect a coin or a nail in parallel to a bulb, the bulb will not light up because current will only flow through the coin or the iron - the bulb is short-circuited.