Physics 9702_s06_qp 2

[XPFMember]

C is shorted! so will take that route...like from X to A to C to Y

 

[DragonCub]

Generally, Short-circuit refers to the case that a part of the circuit becomes in parallel to a zero-resistance conductor (almost zero... In reality it requires harsh condition to achieve exact zero).

Suggest that a lamp in a circuit is get connected parallel to a wire - the wire has ≈0 resistance, then the current would NEVER go through the lamp, and the wire becomes its preferred route. This is short-circuit.
Also you can do a small test: in a circuit with lamps, get an extra wire and connect it DIRECTLY to the two terminals of the power supply. Once connected you will see the lamps all go out. (DON'T KEEP THE WIRE CONNECTED FOR TOO LONG OR THE POWER SUPPLY WILL BE DAMAGED)

In the question, Lamp C is short-circuited, it can then be totally regarded as a wire. So relatively, Lamp B is also short-circuited. Then the B&C part has no resistance. The total R will only be the R of Lamp A.

And for (b) the reason is just the block capital included in the bracket above. Once if Lamp A is faulted (we don't know which is faulted at first) and it is short-circuited, the power supply would be damaged when S1 closed S2 S3 open - here the total R will be 0!

 

[XPFMember]

Assalamoalaikum!!

can u explain part b a bit more...i didnt get you...sorry for that

 

[DragonCub]

That's okay. Actually the short-circuit problem is a comparatively serious issue.

Imagine, if the whole circuit is shorted, that has the same effect as connecting the two terminals of a cell/battery together, what would happen to the cell/battery?
If you have learnt the equation of Electric Power ( P = V * I ) this would be quite understandable.
When the cell/battery is connected directly from +ve terminal to -ve terminal, the total R is actually very very very small. And, according to I = V / R, the current will be very large. While V is still that value, the output power is actually huge.
And then what is gonna receive and use this high rate of energy output? No resistors, no lamps... Totally no electricity-consuming components. The energy would flow back to the cell/battery and work is done just in it.
Electricity energy will all be converted into internal energy (heat) of the cell/battery. The temperature of the cell/battery soars up immediately, causing thermal damage to the cell. Or, under more violent current flow, the cell is unable to endure so huge an energy output rate that it even explodes - that is very hazardous.

So that's why in (b) there is a risk of causing trouble to the power supply if Lamp A short-circuited.

 

[XPFMember]

ok so that means...we are saying that incase if A was shorted then cud damage the power supply ?

Jazak ALlah Khair

 

[DragonCub]

Yeah... Exactly what the question means.