What happens in a parallel circuit when more resistances are added to it? And why?

[diogo_sg]

Hello. Something has been bothering me for a while now:
Consider a circuit with one battery and only one light bulb, for example. We have a value for the resistance of the bulb (R) and there is also one value for voltage (U) and one for current (I).
Now, let's imagine that we connect a new "loop" to this circuit, turning the latter into a parallel circuit. This new loop has yet another bulb (also with resistance R). From KVL, the voltage in this branch of the circuit will be the same as the other branch (U), correct?
But what about current (I')? Will it be the same as the other bulb's (I)? And what will happen to the latter (I)? Will it decrease, increase or remain constant? And why will that happen? Also, what would happen if the second bulb's resistance was bigger or smaller than the first one's (R' > R or R' < R)?

Thank you very much

 

[sophiecentaur]

You know of KVL so I can't understand where your problem lies. The battery is assumed to be ideal (Voltage source which remains the same, whatever current you take) so each possible path between battery terminals stands on its own. Why should connecting a bulb in parallel with the first bulb make any difference to it if it still has the same Potential Difference across it? The bulbs cannot 'know about' each other unless there is a non ideal battery, with some significant series resistance in it, for instance.
You have to remember that simple circuit theory makes assumptions about the components and allows you to do very straightforward Maths calculations. There is no need to have intuitive reservations in simple circuit theory, any more than in simple arithmetic. We believe 2+2 =4. We realize that adding 2 cups of water to 2 cups of water, in a bowl, may not result in exactly 4 cups all arriving in the bowl but we still go along with the simple arithmetic. Same with Basic Electronics.

 

[diogo_sg]

You know of KVL so I can't understand where you problem lies. The battery is assumed to be ideal (Voltage source which remains the same, whatever current you take) so each possible path between battery terminals stands on its own. Why should connecting a bulb in parallel with the first bulb make any difference to it if it still has the same Potential Difference across it? The bulbs cannot 'know about' each other unless there is a non ideal battery, with some significant series resistance in it, for instance.
You have to remember that simple circuit theory makes assumptions about the components and allows you to do very straightforward Maths calculations. There is no need to have intuitive reservations in simple circuit theory, any more than in simple arithmetic. We believe 2+2 =4. We realize that adding 2 cups of water to 2 cups of water, in a bowl, may not result in exactly 4 cups all arriving in the bowl but we still go along with the simple arithmetic. Same with Basic Electronics.

First of all, thank you for the quick reply.
Maybe I should have said that I'm no expert in Kirchhoff's Laws; I just read about them on some physics websites. My apologies.
I understand that, if the bulb's voltage remains the same and its resistance is constant, then the current won't change, despite a new bulb having been added to the circuit.
Although, ever since we start learning about circuits at school, we're told that by adding a new bulb to a circuit (in parallel) the current splits between the several branches, according to the resistance of each of the connected components. So, wouldn't the current change after all?

 

[Logical Dog]

In a parallel circuit, the total equivalent resistance "seen" by the circuit decreases as more resistors are added, irrespective of their value. The parallel resistance is always less than or at most equal to the value of the smallest resistor in the circuit.

The greatest current flows (V/R) to the resistive element with the least value of resistance as voltage is constant when elements are connected in parallel. The current divides between the ratio of total resistance over resistance of one particular branch. (You can derive the current divider rule).

So the smallest resistor also receives the greatest power.

It is very much the "dual" of the series circuit, where in the opposite occurs with resistances, power and voltage. The voltage divider is also the "dual" of the current divider where the ratio of voltages dropped off are in the ratio of the resistor over total.

You can derive kirchhoffs laws through maxwells equations and some highly mathematical manipulations involving something called lumped element models, but they are very neat and easy so we should take them as is :)

I have used Boylestadts circuit analysis book.

 

[Nugatory]

Although, ever since we start learning about circuits at school, we're told that by adding a new bulb to a circuit (in parallel) the current splits between the several branches, according to the resistance of each of the connected components. So, wouldn't the current change after all?

If you have a constant voltage power supply, the current through each loop will be unchanged as you add loops, but of course the total current will increase as you'readding contributions from more loops.

 

[Drakkith]

Although, ever since we start learning about circuits at school, we're told that by adding a new bulb to a circuit (in parallel) the current splits between the several branches, according to the resistance of each of the connected components. So, wouldn't the current change after all?

The current through each branch is determined solely by the voltage applied to that branch and the resistance of the branch. In this example both branches have the same applied voltage and the same resistance. The current through each branch is the same as in the single-branch example, but you have two branches now and so the total current doubles compared to the single-branch example.

 

[Dale]

Although, ever since we start learning about circuits at school, we're told that by adding a new bulb to a circuit (in parallel) the current splits between the several branches, according to the resistance of each of the connected components.

The current does split, but it also increases. It is not a zero sum game. So you get double the current, split in two equal parts (for an identical second resistance)

 

[diogo_sg]

Thank you all for the replies; they were very much appreciated. I now understand what my problem was: I was considering that the total current (the current "produced" by the battery) would remain constant, even with a decrease in the equivalent resistance. I get it now, thanks to you :)

 

[Logical Dog]

my professer said the circuit is "accelerated" when the total resistance seen is decreased or current somehow increases, it makes sense as current is the organised flow of charged particles in a given time period and resistance can be thought as analogous to friction or opposition to motion of charge. so in the branch where the opposition is the least the acceleration is more. So as one decreases the total resistance by adding in parallel each branch is getting accelerated.

 

[Drakkith]

my professer said the circuit is "accelerated" when the total resistance seen is decreased or current somehow increases, it makes sense as current is the organised flow of charged particles in a given time period and resistance can be thought as analogous to friction or opposition to motion of charge. so in the branch where the opposition is the least the acceleration is more. So as one decreases the total resistance by adding in parallel each branch is getting accelerated.

I don't really like your professor's explanation. Thinking about circuits in terms of acceleration is most likely going to end up confusing you more than anything else in my opinion. Current is literally the flow of charge, so adding another branch in parallel to a circuit is like adding a 2nd pipe to a large tank of water. The pressure in both pipes is the same, so the total amount of water flowing out of the tank increases. Changing the size of the 2nd pipe changes the amount of water flowing out through that pipe but does nothing to the 1st pipe.

 

[sophiecentaur]

my professer said the circuit is "accelerated"

I am always amused by the term "professor" that tends to be used instead of "teacher" in some institutions. A 'real' Professor is someone who has a team of post grads under him in a University. If a teacher tells you something then it could well be right. If a real Professor tells you something then it should be right - at least, if it concerns Undergraduate knowledge and below. There are no human beings who can know everything about everything. Even Feynman had his limits -despite what he suggests in his autobiography haha.

 

[Logical Dog]

I am always amused by the term "professor" that tends to be used instead of "teacher" in some institutions. A 'real' Professor is someone who has a team of post grads under him in a University. If a teacher tells you something then it could well be right. If a real Professor tells you something then it should be right - at least, if it concerns Undergraduate knowledge and below. There are no human beings who can know everything about everything. Even Feynman had his limits -despite what he suggests in his autobiography haha.

yes he is a prof, he is amazing and can look at circuits and knows all the available resistors of a certain series and other electronic components by heart. He can see your circuit from miles away and come and correct it. :-) He can caclulate resistances in his head and just by hearing any colour combination you tell him. He assembles circuits in 0.1 of the time it takes us students. He is a real engineer of the highest order (just like some of the seniors here).

 

[sophiecentaur]

yes he is a prof, he is amazing and can look at circuits and knows all the available resistors by heart. He can see your circuit from miles away and come and correct it. :-) He can caclulate resistances in his head and just by hearing any colour combination you tell him.

OK. Well, I suggest you query his use of the term of "accelerate" and ask him what he actually meant was accelerating. Tell hum you know that electrons flow through a wire at an average speed of about 1mm per second. (Look up "electron drift velocity" on Google)
I know a lot of 'humble technicians' who know the resistor colour codes too. Eveything is relative.

 

[Logical Dog]

OK. Well, I suggest you query his use of the term of "accelerate" and ask him what he actually meant was accelerating. Tell hum you know that electrons flow through a wire at an average speed of about 1mm per second. (Look up "electron drift velocity" on Google)
I know a lot of 'humble technicians' who know the resistor colour codes too. Eveything is relative.

Yes, I also know this, and in AC signals electrons dance back and forth, but he was teaching something about a BJT in a switch mode when he mentioned it. He has worked a lot in industry and has some patents to his name too so we all take him very seriously

 

[sophiecentaur]

Not light bulbs?

 

[Logical Dog]

Not light bulbs?

No we have actually never used lightbulbs, we are now dealing with more complicated things. :p
Anyway the prof always laughs at my circuits and my feeble understanding of this field.

 

[sophiecentaur]

No we have actually never used lightbulbs, we are now dealing with more complicated things. :p

Your OP distinctly mentions light bulbs. This is really not helpful. Light bulbs are in fact much more "complicated" than plain resistors. Their resistance changes by a factor of ten from cold to hot.
What level of education are you at?

 

[phinds]

In a parallel circuit, the total equivalent resistance "seen" by the circuit decreases as more resistors are added, irrespective of their value. The parallel resistance is always less than or at most equal to the value of the smallest resistor in the circuit.

Just to be clear, unless the added parallel resistor has infinite resistance (which means that in effect it has not been added) it is impossible for the resultant resistance to be equal to the value of the smallest resistor, it can ONLY be smaller.

 

[sophiecentaur]

1/R₁≤ 1/R₁ + 1/R₂
or
R₁ ≥ R₁ . R₂/(R₁ + R₂)
?
You'd have to wave a lot of hands to put it more precisely than that.