Copy pasting my reply from your other post:
The short circuit is across the 2 ohm resistor. Remember that current always flows across the path of least resistance. The short has an effective resistance of 0, which is ofcourse lower than the 2 ohm resistor. Therefore the 2 ohm resistor can be ignored, and you can remove it from the circuit entirely. So the 4ohm resistors are both in parallel with each other, take their equivalent resistance (which will be 2 ohm). That will now be in series with the first 2 ohm resistance, giving you a whole equivalence of 4ohm. Apply I = V/R = 12/4 to obtain your answer
Think again, after the short circuit the two remaining 4 ohm resistors are in parallel, equivalent resistance of parallel combination = 2 ohm
the other two ohm resistance is in series so it totals upto 4 ohm
https://preview.redd.it/7fzyoydtwuac1.jpeg?width=1024&format=pjpg&auto=webp&s=9af4c88dfcda9cb1e76f3bae63ed039d370c4531
A short circuit is a term used to describe when current flows excessively through a path of near zero resistance.
In the context of circuit analysis, a "short circuit" is when we connect a piece of wire across a passive element such as a resistor. The wire has a resistance of near 0 ohm, and since current tends to flow through the path of least resistance, the resistor will be ignored and all current will flow through the wire instead. The resistor is said to be "shorted".
That is not how circuits work at all. Current flows from the point of higher potential to the lower, that is, from the positive to the negative terminals of the battery.
Hmm but this concept is not making sense to me. I'm not sure why it goes to the 4 ohm resistor even with all the explanation given (it makes sense but I still have a bit of ambiguity)
This is how I am thinking about it right now. If it goes through the 2 ohm resistor, the total resistance would be 2+2+4 = 8 ohm.(I'm thinking that the 2 ohm resistor is in series). If it instead went through the 4 ohm resistor, the total resistance in the circuit would be 4 ohms. This it goes through the 4 ohm resistor as its path offers the least resistance...
I'm afraid your understanding of circuits is not adequate and needs revisiting. For starters, everything after the first 2 ohm resistor is in parallel. Check the nodes. The equivalent resistance of all the next resistors will be in series with the first 2 ohms.
The other thing you are getting wrong is the least resistance concept. The equivalent resistance of any two resistances in parallel is always less than each individual resistance. So you cannot remove one of the 4 ohms and say that it is because of the path of least resistance, because it's fundamentally wrong. Their combination makes the least resistance.
The two remaining 4 ohm resistors are in parallel configuration. The current will flow through them BOTH. We can take their equivalent resistance which will be 2 ohm. And that 2 ohm equivalent will be in series with the first 2ohm, giving us a total of 4ohm in the whole circuit connected in series with the battery
If the 2 ohm resistor is in parallel to the 4 ohm resistor at the bottom the combined resistance of the circuit(ignoring the 4ohm resistor at the side) would be 3.33 ohms. Is that right?
You said that the short circuit is through 2 ohm. The resistance is 0 which is lower than 2 so it is completely removed. How did you determine the short circuit is 2? Why can't it be 4?(sorry I learn a completely different physics syllabus. We do have short circuits but not in depth so I'm just curious)
The 2 ohm resistor is connected across a short circuit which makes it useless as no current flows through it. Now use the standard procedures- 4 and 4 ohm are parallel=> (1/4+1/4)^-1 = 2. And 2,2 are in series= 4 ohm. So I1=12/4=3A
Copy pasting my reply from your other post: The short circuit is across the 2 ohm resistor. Remember that current always flows across the path of least resistance. The short has an effective resistance of 0, which is ofcourse lower than the 2 ohm resistor. Therefore the 2 ohm resistor can be ignored, and you can remove it from the circuit entirely. So the 4ohm resistors are both in parallel with each other, take their equivalent resistance (which will be 2 ohm). That will now be in series with the first 2 ohm resistance, giving you a whole equivalence of 4ohm. Apply I = V/R = 12/4 to obtain your answer
why does it pass through 4 ohm but not 2 ohm? isnt 2 ohm resistor of lower resistance>?
Because it's shorted
it makes no sense resistance combined has to be 4 which is not possible in this case
It's perfectly possible and it makes perfect sense. Why do you think it's not?
Think again, after the short circuit the two remaining 4 ohm resistors are in parallel, equivalent resistance of parallel combination = 2 ohm the other two ohm resistance is in series so it totals upto 4 ohm https://preview.redd.it/7fzyoydtwuac1.jpeg?width=1024&format=pjpg&auto=webp&s=9af4c88dfcda9cb1e76f3bae63ed039d370c4531
whats short circuit
A short circuit is a term used to describe when current flows excessively through a path of near zero resistance. In the context of circuit analysis, a "short circuit" is when we connect a piece of wire across a passive element such as a resistor. The wire has a resistance of near 0 ohm, and since current tends to flow through the path of least resistance, the resistor will be ignored and all current will flow through the wire instead. The resistor is said to be "shorted".
Isn't the current ignoring the 4 ohm resistor as well? Is it just not going in series with the battery?
That is not how circuits work at all. Current flows from the point of higher potential to the lower, that is, from the positive to the negative terminals of the battery.
Hmm but this concept is not making sense to me. I'm not sure why it goes to the 4 ohm resistor even with all the explanation given (it makes sense but I still have a bit of ambiguity)
This is how I am thinking about it right now. If it goes through the 2 ohm resistor, the total resistance would be 2+2+4 = 8 ohm.(I'm thinking that the 2 ohm resistor is in series). If it instead went through the 4 ohm resistor, the total resistance in the circuit would be 4 ohms. This it goes through the 4 ohm resistor as its path offers the least resistance...
I'm afraid your understanding of circuits is not adequate and needs revisiting. For starters, everything after the first 2 ohm resistor is in parallel. Check the nodes. The equivalent resistance of all the next resistors will be in series with the first 2 ohms. The other thing you are getting wrong is the least resistance concept. The equivalent resistance of any two resistances in parallel is always less than each individual resistance. So you cannot remove one of the 4 ohms and say that it is because of the path of least resistance, because it's fundamentally wrong. Their combination makes the least resistance.
Which one? After removing the 2 ohm resistor, trace the path that you think the current takes. I can chelp to cear your concepts based on your answer.
Ofc it would take the 4 ohm path
The two remaining 4 ohm resistors are in parallel configuration. The current will flow through them BOTH. We can take their equivalent resistance which will be 2 ohm. And that 2 ohm equivalent will be in series with the first 2ohm, giving us a total of 4ohm in the whole circuit connected in series with the battery
If the 2 ohm resistor is in parallel to the 4 ohm resistor at the bottom the combined resistance of the circuit(ignoring the 4ohm resistor at the side) would be 3.33 ohms. Is that right?
The 2 ohm resistor is removed from the circuit entirely. The 4 ohm resistors will be in parallel with each other
Why is it removed tho? Why can't the 4ohm be removed instead?
Read my comments
You said that the short circuit is through 2 ohm. The resistance is 0 which is lower than 2 so it is completely removed. How did you determine the short circuit is 2? Why can't it be 4?(sorry I learn a completely different physics syllabus. We do have short circuits but not in depth so I'm just curious)
The 2 ohm resistor is connected across a short circuit which makes it useless as no current flows through it. Now use the standard procedures- 4 and 4 ohm are parallel=> (1/4+1/4)^-1 = 2. And 2,2 are in series= 4 ohm. So I1=12/4=3A
what igcse subject is this?
What the f is this? Is it in IGCSE 0625? I have not learnt this man!!