Original post by As.1997
Can someone please explain this — I really do not know what is going on.
Original post by As.1997
(i) Identical cells connected in parallel would NOT change the overall voltage across the cell. Make sure you read the statement carefully: it is identical cells that are connected in parallel.
https://www.primaryconnections.org.au/sites/all/modules/primaryconnections/includes/SBR/data/Phy/sub/seriespara/seriespara.htm
(ii) Since the overall voltage measured by the voltmeter does not change, the current will not change, too.
(iii) I assume the bulbs are identical. You can plug in some values (emf = 2 V, the resistance of the bulb = 2 Ω) to perform the calculation of current before and after S2 is closed.
(iv) Once the calculation is done, you would know the answer.
(v) When S3 is closed, there is short-circuit.
Original post by As.1997
Would really appreciate if someone could explain this to me 
This is just a day old.
Original post by Eimmanuel
(i) Identical cells connected in parallel would NOT change the overall voltage across the cell. Make sure you read the statement carefully: it is identical cells that are connected in parallel.
https://www.primaryconnections.org.au/sites/all/modules/primaryconnections/includes/SBR/data/Phy/sub/seriespara/seriespara.htm
(ii) Since the overall voltage measured by the voltmeter does not change, the current will not change, too.
(iii) I assume the bulbs are identical. You can plug in some values (emf = 2 V, the resistance of the bulb = 2 Ω) to perform the calculation of current before and after S2 is closed.
(iv) Once the calculation is done, you would know the answer.
(v) When S3 is closed, there is short-circuit.
This is just a day old.
https://www.primaryconnections.org.au/sites/all/modules/primaryconnections/includes/SBR/data/Phy/sub/seriespara/seriespara.htm
(ii) Since the overall voltage measured by the voltmeter does not change, the current will not change, too.
(iii) I assume the bulbs are identical. You can plug in some values (emf = 2 V, the resistance of the bulb = 2 Ω) to perform the calculation of current before and after S2 is closed.
(iv) Once the calculation is done, you would know the answer.
(v) When S3 is closed, there is short-circuit.
This is just a day old.
For iii) Can't you logically assume that that the reading on A2 will increase since closing S2 will create a parallel circuit which will mean the overall current should increase because of a decrease in total resistance? This would, therefore, make it unnecessary to perform any theoretical calculations? (I'm just assuming, please correct me if I'm wrong -- my basics for electricity is poor so I'm just learning as I practice questions)
For iv) Again for this can't you logically assume that the current will increase in A3 since a parallel circuit is produced which will mean a lower total resistance and therefore a greater current flow?
Also, the website mentions,
"A 9 Volt battery will produce a voltage 6 times larger than a single 1.5 Volt battery in the same circuit, but the current in each circuit will be the same no matter where the current is measured.
This happens because the batteries are arranged in a line, and like water flowing through different hoses connected in a line, what goes in one end must come out the other. The same electrons must flow through all the batteries at the same rate, so the current must be the same in each battery and in each part of the circuit."
MY QUESTION:
If the current in the 9V and 1.5V circuits are the same (i.e. lets say 2Amps), then in this equation V=IR, did the resistance increase when we used the 9V battery to make up for the increase in the voltage?
"A 9 Volt battery will produce a voltage 6 times larger than a single 1.5 Volt battery in the same circuit, but the current in each circuit will be the same no matter where the current is measured.
This happens because the batteries are arranged in a line, and like water flowing through different hoses connected in a line, what goes in one end must come out the other. The same electrons must flow through all the batteries at the same rate, so the current must be the same in each battery and in each part of the circuit."
MY QUESTION:
If the current in the 9V and 1.5V circuits are the same (i.e. lets say 2Amps), then in this equation V=IR, did the resistance increase when we used the 9V battery to make up for the increase in the voltage?
(edited 4 years ago)
Original post by As.1997
Also, the website mentions,
"A 9 Volt battery will produce a voltage 6 times larger than a single 1.5 Volt battery in the same circuit, but the current in each circuit will be the same no matter where the current is measured.
This happens because the batteries are arranged in a line, and like water flowing through different hoses connected in a line, what goes in one end must come out the other. The same electrons must flow through all the batteries at the same rate, so the current must be the same in each battery and in each part of the circuit."
MY QUESTION:
If the current in the 9V and 1.5V circuits are the same (i.e. lets say 2Amps), then in this equation V=IR, did the resistance increase when we used to 9V battery to make up for the increase in the voltage?
"A 9 Volt battery will produce a voltage 6 times larger than a single 1.5 Volt battery in the same circuit, but the current in each circuit will be the same no matter where the current is measured.
This happens because the batteries are arranged in a line, and like water flowing through different hoses connected in a line, what goes in one end must come out the other. The same electrons must flow through all the batteries at the same rate, so the current must be the same in each battery and in each part of the circuit."
MY QUESTION:
If the current in the 9V and 1.5V circuits are the same (i.e. lets say 2Amps), then in this equation V=IR, did the resistance increase when we used to 9V battery to make up for the increase in the voltage?
Dunno exactly what the books saying tbh. Is there a diagram or some more context?
Would imagine it's saying that the current is the same at different parts of the series circuit when using a 1.5 V battery and that the current is the same at different parts of the series circuit when using a 9V battery... This is how series circuits work.
But that's not the same as saying the current is the same when using either a 1.5V battery or a 9V battery
Original post by Joinedup
Dunno exactly what the books saying tbh. Is there a diagram or some more context?
Would imagine it's saying that the current is the same at different parts of the series circuit when using a 1.5 V battery and that the current is the same at different parts of the series circuit when using a 9V battery... This is how series circuits work.
But that's not the same as saying the current is the same when using either a 1.5V battery or a 9V battery
Would imagine it's saying that the current is the same at different parts of the series circuit when using a 1.5 V battery and that the current is the same at different parts of the series circuit when using a 9V battery... This is how series circuits work.
But that's not the same as saying the current is the same when using either a 1.5V battery or a 9V battery
Oh I see I think that’s what they were trying to say
Original post by As.1997
For iii) Can't you logically assume that that the reading on A2 will increase since closing S2 will create a parallel circuit which will mean the overall current should increase because of a decrease in total resistance? This would, therefore, make it unnecessary to perform any theoretical calculations? (I'm just assuming, please correct me if I'm wrong -- my basics for electricity is poor so I'm just learning as I practice questions) …
You can definitely choose NOT to do the calculation if you can explain the effect correctly. Your reasoning of the decrease in the total effective resistance will bring about the increase in the total current in A2 is good.
Original post by As.1997
….For iv) Again for this can't you logically assume that the current will increase in A3 since a parallel circuit is produced which will mean a lower total resistance and therefore a greater current flow?
The increase of total current in A2 does not mean that the current in A3 would increase. If you have done the calculation, you should realize that the current in A3 remains the same. Why? I would leave it to you to find out the answer.
Original post by As.1997
Also, the website mentions,
"A 9 Volt battery will produce a voltage 6 times larger than a single 1.5 Volt battery in the same circuit, but the current in each circuit will be the same no matter where the current is measured.
This happens because the batteries are arranged in a line, and like water flowing through different hoses connected in a line, what goes in one end must come out the other. The same electrons must flow through all the batteries at the same rate, so the current must be the same in each battery and in each part of the circuit."
MY QUESTION:
If the current in the 9V and 1.5V circuits are the same (i.e. lets say 2Amps), then in this equation V=IR, did the resistance increase when we used the 9V battery to make up for the increase in the voltage?
"A 9 Volt battery will produce a voltage 6 times larger than a single 1.5 Volt battery in the same circuit, but the current in each circuit will be the same no matter where the current is measured.
This happens because the batteries are arranged in a line, and like water flowing through different hoses connected in a line, what goes in one end must come out the other. The same electrons must flow through all the batteries at the same rate, so the current must be the same in each battery and in each part of the circuit."
MY QUESTION:
If the current in the 9V and 1.5V circuits are the same (i.e. lets say 2Amps), then in this equation V=IR, did the resistance increase when we used the 9V battery to make up for the increase in the voltage?
I link the website is to let you know the given conclusion. Some of the writing on the website is not precise and also not really correct. I forget to tell you that I want you to look at the conclusion and NOT warning you the caveats on the website.
The description that you quoted is one of the imprecise writing: what it “really” want to explain (where I interpret based on what is given before and after) is to compare the circuit with a single 9 V cell and a circuit that has six 1.5 V cells connected in series (that gives the total voltage to be 9 V).
As long as you know what happens when identical cells are connected in series and parallel, respectively, you are good.
If S3 is closed, the overall resistance decreases since parallel circuit is formed. Does this mean that the reading in A3 increases?
If so is there a way to work it out? (I’ve tried to work it out below)
Let’s say the resistance for the bulbs are 2ohms and the bulbs are identical. At the same time let’s say the cell has a voltage of 6V. Prior to closing S2 this would mean the current is 3Amps (6/2=3)
After closing S2 the total resistance is 1ohm. (1/Rt =1/2 1/2, therefore Rt is 1ohm)
Therefore the total current is 6/1=6Amps after closing S2. Does this mean that the loop containing A3 will have 3Amps and the loop below with just a bulb also having 3Amps?
If so is there a way to work it out? (I’ve tried to work it out below)
Let’s say the resistance for the bulbs are 2ohms and the bulbs are identical. At the same time let’s say the cell has a voltage of 6V. Prior to closing S2 this would mean the current is 3Amps (6/2=3)
After closing S2 the total resistance is 1ohm. (1/Rt =1/2 1/2, therefore Rt is 1ohm)
Therefore the total current is 6/1=6Amps after closing S2. Does this mean that the loop containing A3 will have 3Amps and the loop below with just a bulb also having 3Amps?
(edited 4 years ago)
Original post by As.1997
If S3 is closed, the overall resistance decreases since parallel circuit is formed. Does this mean that the reading in A3 increases?
If so is there a way to work it out? (I’ve tried to work it out below)
Let’s say the resistance for the bulbs are 2ohms and the bulbs are identical. At the same time let’s say the cell has a voltage of 6V. Prior to closing S2 this would mean the current is 3Amps (6/2=3)
After closing S2 the total resistance is 1ohm. (1/Rt =1/2 1/2, therefore Rt is 1ohm)
Therefore the total current is 6/1=6Amps after closing S2. Does this mean that the loop containing A3 will have 3Amps and the loop below with just a bulb also having 3Amps?
If so is there a way to work it out? (I’ve tried to work it out below)
Let’s say the resistance for the bulbs are 2ohms and the bulbs are identical. At the same time let’s say the cell has a voltage of 6V. Prior to closing S2 this would mean the current is 3Amps (6/2=3)
After closing S2 the total resistance is 1ohm. (1/Rt =1/2 1/2, therefore Rt is 1ohm)
Therefore the total current is 6/1=6Amps after closing S2. Does this mean that the loop containing A3 will have 3Amps and the loop below with just a bulb also having 3Amps?
A2 measures the total current in the circuit but A2 measures current in the light bulb next to it.
Therefore the total current is 6/1=6Amps after closing S2. Does this mean that the loop containing A3 will have 3Amps and the loop below with just a bulb also having 3Amps?
This is correct which I already mentioned previously. But why? This is what I am asking.
Original post by Eimmanuel
The increase of total current in A2 does not mean that the current in A3 would increase. If you have done the calculation, you should realize that the current in A3 remains the same. Why? I would leave it to you to find out the answer.
Okay, I’ll give it another shot: the reason why iv) is wrong is because the current doesn’t decrease or increase in A3 rather it would stay the same and for this reason it is false. The reason it stays the same is because regardless of whether the bulb is in parallel or series the resistance of the bulb containing A3 doesn’t change (ie it would remain 2ohms) and thus the current would remain 3Amps since 6V\2ohms= 3Amp.
Original post by As.1997
Okay, I’ll give it another shot: the reason why iv) is wrong is because the current doesn’t decrease or increase in A3 rather it would stay the same and for this reason it is false. The reason it stays the same is because regardless of whether the bulb is in parallel or series the resistance of the bulb containing A3 doesn’t change (ie it would remain 2ohms) and thus the current would remain 3Amps since 6V\2ohms= 3Amp.
The current in the light bulb does not change is mainly because the p.d. across the light bulb does not change NOT just the resistance of the light bulb does not change.
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