Voltage and current in a series circuit

[nsv23]

Consider the circuit in the attachment

I know that the current is the same in a series circuit and voltage is divided at the 3 resistors. And resistance is added.
R = R1 + R2 + R3 = 100 + 200 + 300 = 600 Ω.
V = IR, 12 = I * 600, I = 0.02 amps = current flowing through the circuit.
V1 (voltage @ R1) = IR1 = .02 * 100 = 2 v
V2 (voltage @ R2) = IR2 = .02 * 200 = 4 v
V3 (voltage @ R3) = IR3 = .02 * 300 = 6 v
V1 + V2 + V3 = 2 + 4 + 6 = 12v same as the supplied voltage

If I measure the current at point A will it 0.02 A? I mean at A it has not passed through the resistance. So shouldn’t it be greater than 0.02 A and when meeting the resistance it reduces to 0.02A?
Assuming that the LED requires minimum 0.01 amps to glow, out of the 0.02 amps flowing in the circuit will the LED take only 0.01 amps and leave the remaining? If so after passing through the LED the current that flows to the -ve terminal is only 0.01?
When someone says “resistance is a circuit” does it mean that the resistance is there from the beginning of the circuit i.e. from the +ve terminal all the way to through the circuit to –ve terminal? If so how come? In the above case assuming I connect the +ve terminal of the 12V battery to first resistor by a piece of wire, shouldn’t current between +ve terminal and first resistor be greater than the current at the resistor?
Why is their voltage drop and not current reduction? How can voltage drop? Current is the flow of electrons right? When meeting with resistance isn’t the flow of electrons that is being restricted? And not the voltage which is the ‘force’ that ‘pushes’ the electrons?
What exactly happens when the flow of electrons meet with a resistor? Will some electrons be left at the resistor i.e. blocked by resistors and others carry on? For example current before meeting resistance is 10amps and after meeting a resistance reduces to 8amps. So 2amps are left at the resistance in other words electrons carrying the 2 amps are left at the resistance right? What happens to those electrons and current left at resistance?

 

[tiny-tim]

welcome to pf!

hi nsv23! welcome to pf!

Consider the circuit in the attachment

I know that the current is the same in a series circuit and voltage is divided at the 3 resistors. And resistance is added.
R = R1 + R2 + R3 = 100 + 200 + 300 = 600 Ω.
V = IR, 12 = I * 600, I = 0.02 amps = current flowing through the circuit.
V1 (voltage @ R1) = IR1 = .02 * 100 = 2 v
V2 (voltage @ R2) = IR2 = .02 * 200 = 4 v
V3 (voltage @ R3) = IR3 = .02 * 300 = 6 v
V1 + V2 + V3 = 2 + 4 + 6 = 12v same as the supplied voltage

If I measure the current at point A will it 0.02 A?

yes

I mean at A it has not passed through the resistance. So shouldn’t it be greater than 0.02 A and when meeting the resistance it reduces to 0.02A?

no, the current is the same throughout the circuit

Why is their voltage drop and not current reduction? How can voltage drop? Current is the flow of electrons right? When meeting with resistance isn’t the flow of electrons that is being restricted? And not the voltage which is the ‘force’ that ‘pushes’ the electrons?

it's exactly like water

suppose you have a number of water-wheels on a river, and a pump at the start of the river, putting the water under pressure P

as the river goes past each water-wheel, the pressure becomes less and less

but the current stays the same … how can it not do so? … water can't be created or destroyed, and the amount of water passing each point per second must be the same

similarly, electric current is charge per second, charge can't be created or destroyed, so how could the electric current not be the same? ​

Assuming that the LED requires minimum 0.01 amps to glow, out of the 0.02 amps flowing in the circuit will the LED take only 0.01 amps and leave the remaining? If so after passing through the LED the current that flows to the -ve terminal is only 0.01?

if it needs 0.01 A, then it's getting twice as much, and it will probably glow twice as bright (or melt)

When someone says “resistance is a circuit” does it mean that the resistance is there from the beginning of the circuit i.e. from the +ve terminal all the way to through the circuit to –ve terminal?

not exactly

it's like pushing a lot of blocks, you only push the back block, but you have to push hard enough to push all of them!

(mass is resistance to force, electric resistance is resistance to electric force)

In the above case assuming I connect the +ve terminal of the 12V battery to first resistor by a piece of wire, shouldn’t current between +ve terminal and first resistor be greater than the current at the resistor?

i don't understand

What exactly happens when the flow of electrons meet with a resistor? Will some electrons be left at the resistor i.e. blocked by resistors and others carry on? For example current before meeting resistance is 10amps and after meeting a resistance reduces to 8amps. So 2amps are left at the resistance in other words electrons carrying the 2 amps are left at the resistance right? What happens to those electrons and current left at resistance?

no electrons are left behind, if they were, then there would be a build-up of negative charge (only happens at a capacitor, not a resistor)

the current is electrons per second, and that's the same everywhere

 

[nsv23]

Thanks for the reply.
"If I measure the current at point A will it 0.02 A? I mean at A it has not passed through the resistance. So shouldn’t it be greater than 0.02 A and when meeting the resistance it reduces to 0.02A?"

no, the current is the same throughout the circuit

I still have doubts about it. I am posting the same question in a different way.

"In the above case assuming I connect the +ve terminal of the 12V battery to first resistor by a piece of wire, shouldn’t current between +ve terminal and first resistor be greater than the current at the resistor?"

i don't understand

Both the questions are related. Let me elaborate.
Consider the first circuit. Here a wire is between the resistor and the battery’s +ve terminal and the wire is connecting them. Now current has to flow through the wire and reach R1 right? Between the +ve terminal and R1 there is no resistance except for the resistance of the wire which is very low and negligible. So therefore shouldn’t current be high in the wire and at the point of contact with R1 shouldn’t it reduce to 0.02 amps?
Now consider the second circuit below. R1 is directly connected to the +ve terminal of the battery. So as soon as the circuit is closed, the current flowing from the +ve terminal meets with resistance and is reduced to 0.02 amps, right?

"Assuming that the LED requires minimum 0.01 amps to glow, out of the 0.02 amps flowing in the circuit will the LED take only 0.01 amps and leave the remaining? If so after passing through the LED the current that flows to the -ve terminal is only 0.01?"

if it needs 0.01 A, then it's getting twice as much, and it will probably glow twice as bright (or melt)

Why does it take all of 0.02 amps? Assuming the LED's maximum is 0.01 amps. Why does it melt, I mean can't it just accept 0.01 amps and leave the rest? If it leaves the rest isn’t the remaining 0.01 amps flowing back to the –ve terminal? Which means the current is not the same everywhere.
Now assuming LED's maximum is 0.02 amps, when current reaches the LED, it glows because it has absorbed the 0.02amps? If so there shouldn't be any current flowing after the LED right? Or taking the Water analogy, a tank filled with water, at one end there is water wheel. What happens is the pressure (voltage) causes the water (current) to flow and hit the water wheel (LED) which rotates (glows). After hitting the water wheel the water flows into a pipe which is connected to back tank. Is this what happens: because of voltage, current flows from +ve terminal, reaches the LED, lights it, goes back to the -ve terminal? The cycle repeats as long as the circuit is closed. Why does the LED glow continuously? I mean after hitting the water wheel it rotates and shouldn’t it stop? It is not stopping because of the speed at which the current flows? i.e. current lights the LED, leaves it, goes back to battery and starts again from the +ve terminal, all this in fraction of seconds, that the LED is on all the time. Or, is it because the flow is continuous, that the LED is getting constant supply and not because of speed. Like water from tank, the front of water hits the wheel and goes into the pipe and since there is still the back of water in the tank as the front of the water goes into the tank there is water coming out of it. I hope it is not confusing

 

[tiny-tim]

hi nsv23!

"In the above case assuming I connect the +ve terminal of the 12V battery to first resistor by a piece of wire, shouldn’t current between +ve terminal and first resistor be greater than the current at the resistor?"

no, the current is the same throughout the circuit

Consider the first circuit …
Now consider the second circuit …

they're the same

Why does it take all of 0.02 amps? Assuming the LED's maximum is 0.01 amps. Why does it melt, I mean can't it just accept 0.01 amps and leave the rest?

it has no option

1 amp = 1 coulomb per second = 6.242*10¹⁸ electrons per second

0.01 amp is 6.242*10¹⁶ electrons per second

that's what amps mean!

there's 6.242*10¹⁶ electrons going through the resistor, and through the wire, and through everything else in the circuit, every second

how can the resistor ignore some of them?

if a water-wheel is supposed to operate at 100 gallons per second, and you let 200 gallons per second go through the water channel, then it will either go round twice as fast, or it will break

the only way the water-wheel can "leave the rest" is to have another parallel channel (or lift the water-wheel partly out, which is the same thing), exactly like putting another resistor in parallel

 

[sardar]

I can confirm that your understanding of voltage and current in a series circuit is correct. In a series circuit, the current remains constant throughout the circuit, while the voltage is divided among the various components. The total resistance in a series circuit is the sum of the individual resistances, as shown in your calculation.

To answer your questions, yes, the current at point A will be 0.02 amps, as the current is the same at all points in a series circuit. The concept of voltage drop is based on Ohm's Law, which states that the voltage drop across a resistor is directly proportional to the current flowing through it. So, even though the current remains constant, the voltage drop will vary at different points in the circuit, depending on the resistance.

In the case of the LED, it will indeed take only 0.01 amps, as it has a higher resistance compared to the other resistors in the circuit. The remaining 0.01 amps will continue to flow through the rest of the circuit.

Resistance is present throughout the entire circuit, from the positive terminal to the negative terminal. This is because all materials have some level of resistance, and as the current flows through them, there will be a voltage drop. In your example, if you connect the positive terminal of the battery directly to the first resistor, the current will be the same at both points, but the voltage drop will be higher at the resistor compared to the wire.

When the flow of electrons meets a resistor, it encounters resistance in the form of collisions with atoms in the material. This resistance impedes the flow of electrons, resulting in a voltage drop. Some electrons may be left at the resistor, but they will eventually continue to flow through the circuit. The current left at the resistor is not "blocked," but rather, it is a result of the voltage drop and the resistance of the material.

In summary, voltage and current in a series circuit are interconnected, and their values can vary at different points in the circuit due to the presence of resistance. The flow of electrons is affected by the resistance, resulting in voltage drops and variations in current. I hope this helps clarify your understanding of voltage and current in a series circuit.