Clearing up confusion on Resistors and Current in Parallel Circuits

[betterfuture]

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Adding a resistor in parallel decreases the overall Resistance and increases the total Current of the circuit. However, where do all these extra charge come from? I know the battery provides the potential difference to drive the flow of electrons to do work, but where did the extra charges come from when the current increased? Did more electrons suddenly became available as soon as the resistor was added in parallel?

 

[NextStepTutor_3]

Current is measured in amperes, or coulombs per second! So an increase in current doesn't necessarily mean that more coulombs of charge have suddenly been introduced. It more commonly means that charge is moving faster (in other words, it takes fewer seconds for the same amount of charge to travel).

 

[betterfuture]

If you look at the equation for current, I=q/t, an increase in charge would be increase in current, correct?

So when you add a resistor in parallel, the resistors are added to get the total resistance by 1/R1+ 1/R2 = Req. So when a resistor is added, doesn't the total current increase? Where is this extra increase in current coming from? Could you explain please?

 

[NextStepTutor_3]

Yes, an increase in charge does produce an increase in current (when time is constant)! But look at that equation, I = q/t. A decrease in time also produces an increase in current, even when charge has not changed at all. That's what happens when you add a parallel resistor. Think of it this way:

Adding a parallel resistor is like adding an extra branch that charges can travel through. No matter what (even if this resistor has a very high resistance), adding an alternative path must make it easier for charge to move overall. If this resistor has a low resistance, it'll make it much easier; if it has a high resistance, it'll only make it slightly easier, but the effective resistance of the circuit will decrease either way.

Now, what does it mean to have a low resistance? To put it very simply, it means that charges can move more rapidly through the material. Let's say that one particular charge previously took 10 ms to move through the circuit, and now it takes 8 ms. Our "time" value is smaller, which makes current increase. No "extra charges" are appearing when we add that resistor - the charges that we have are simply moving faster.

 

[betterfuture]

Okay. Still slightly confused. Hope you can still help.

When resistors are connected in parallel, the voltage drop associated with a resistor does not change because every voltage drop of every resistor is equal to the voltage source. Okay. Got that. Also, the resistors that are connected in parallel have a RESISTANCE which, of course, does not change. So only the current running through each resistor is different, but again, the sum of each current equals the total current of the circuit. Okay. Got that as well.

Now, say, I added another resistor in parallel. Adding this resistor,
A) Does NOT change the others resistors resistance value
B) Does NOT change the voltage drop associated with the other resistors, because the voltage drop is equal to the voltage source
C) Also, Does NOT change the current running through the other resistors because neither the voltage drop nor the other resistors resistance value has changed, so current doesn't change. It stays the same.

Then that means the current for the new resistor could be calculated from V=IR to get the current running through that resistor.

So if current before was 50 Amps, let's say for example, and now the new resistor is added, the current would obviously increase from 50 Amps to a higher value. Now here is where I get confused. The current running through the old resistors DID NOT CHANGE, but OVERALL CURRENT DID INCREASE. How, now, does time fit into all of this? I thought maybe the increase in current came from the battery allowing more charges to pass through? I am not understanding how time has anything to do with the increase in current.

 

[aldol16]

So if current before was 50 Amps, let's say for example, and now the new resistor is added, the current would obviously increase from 50 Amps to a higher value. Now here is where I get confused. The current running through the old resistors DID NOT CHANGE, but OVERALL CURRENT DID INCREASE. How, now, does time fit into all of this? I thought maybe the increase in current came from the battery allowing more charges to pass through? I am not understanding how time has anything to do with the increase in current.

Overall resistance did change. Overall resistance decreased. Which makes current increase for a given potential difference. You're confusing individual resistances with the overall resistance of the circuit.

Just imagine a river flowing downhill. The downhill gradient is always the same. But you could always make the river run faster or slower by modulating the resistance it faces.

 

[NextStepTutor_3]

I see the issue! @aldol16 is right as usual. Your mistake is assuming that the current moving through the other resistors did not change. When you add a parallel resistor, the resistance of the overall circuit decreases and the current increases. When calculating current, always find the total ("equivalent") resistance of the entire circuit, then use V = IR to determine the current that is exiting the battery. As you do this, you'll see that adding either a parallel or series resistor will change this value for total resistance, and thus change the current. Once you've found the current produced by the battery, you can use Kirchoff's junction rule to find the current that travels through each resistor. Again, the current moving through the "old" resistors will be different after addition of the new one.

 

[betterfuture]

I see the issue! @aldol16 is right as usual. Your mistake is assuming that the current moving through the other resistors did not change. When you add a parallel resistor, the resistance of the overall circuit decreases and the current increases. When calculating current, always find the total ("equivalent") resistance of the entire circuit, then use V = IR to determine the current that is exiting the battery. As you do this, you'll see that adding either a parallel or series resistor will change this value for total resistance, and thus change the current. Once you've found the current produced by the battery, you can use Kirchoff's junction rule to find the current that travels through each resistor. Again, the current moving through the "old" resistors will be different after addition of the new one.

Ohh okay I see now. So basically, after adding the resistor, the overall current changed and I did not take that into account and assumed the current running through the old resistors would not change. Thanks! I appreciate your help. Made it clear as day now.