eli5 How is current the same in a series circuit, if voltage changes?

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I watched so many youtube videos about electronics but i cant seem to understand what is voltage and current. If I = V/R, then in a series circuit, if the voltage is reducing after each component, then the current should also decrease right? I don’t understand the water pipe analogy

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12 Answers

Anonymous 0 Comments

You are not considering the voltage correctly.

You measure voltage between 2 points. So if you got 3 things in series with all the same resistance, A-B-C. Let say we got 6v. It would be 6v A 4v B 2v C 0v in the series.

The voltage between each component is 2v. All the components have the same R so I should be the same through all of them.

The Voltage on A is 6v. The voltage after C is 0v, that a 6v different, 3x bigger, but now you add A+B+C resistance since they are in series, so 3x the voltage over 3x the resistance has you end up at the same current.

Anonymous 0 Comments

You are not considering the voltage correctly.

You measure voltage between 2 points. So if you got 3 things in series with all the same resistance, A-B-C. Let say we got 6v. It would be 6v A 4v B 2v C 0v in the series.

The voltage between each component is 2v. All the components have the same R so I should be the same through all of them.

The Voltage on A is 6v. The voltage after C is 0v, that a 6v different, 3x bigger, but now you add A+B+C resistance since they are in series, so 3x the voltage over 3x the resistance has you end up at the same current.

Anonymous 0 Comments

You are not considering the voltage correctly.

You measure voltage between 2 points. So if you got 3 things in series with all the same resistance, A-B-C. Let say we got 6v. It would be 6v A 4v B 2v C 0v in the series.

The voltage between each component is 2v. All the components have the same R so I should be the same through all of them.

The Voltage on A is 6v. The voltage after C is 0v, that a 6v different, 3x bigger, but now you add A+B+C resistance since they are in series, so 3x the voltage over 3x the resistance has you end up at the same current.

Anonymous 0 Comments

Electrons going down the voltage gradient are like if bumper cars were lined up bumper to bumper pushing down a long downhill ramp.

The car at the top of the hill has the most potential energy; the car at the bottom of the hill has the least. But still, cars **have** to come off the bottom of the ramp as quickly as they are added to the top of the ramp.

The ramp is a series circuit. The cars are electric charges. The rate of traffic flow is current.

Capacitors change the rules a bit, adding a capacitor is like placing a rest stop next to the road, but there’s no way back out except by driving backwards up the offramp.

All of this stuff is required by conservation of mass & energy.

Anonymous 0 Comments

Electrons going down the voltage gradient are like if bumper cars were lined up bumper to bumper pushing down a long downhill ramp.

The car at the top of the hill has the most potential energy; the car at the bottom of the hill has the least. But still, cars **have** to come off the bottom of the ramp as quickly as they are added to the top of the ramp.

The ramp is a series circuit. The cars are electric charges. The rate of traffic flow is current.

Capacitors change the rules a bit, adding a capacitor is like placing a rest stop next to the road, but there’s no way back out except by driving backwards up the offramp.

All of this stuff is required by conservation of mass & energy.

Anonymous 0 Comments

Electrons going down the voltage gradient are like if bumper cars were lined up bumper to bumper pushing down a long downhill ramp.

The car at the top of the hill has the most potential energy; the car at the bottom of the hill has the least. But still, cars **have** to come off the bottom of the ramp as quickly as they are added to the top of the ramp.

The ramp is a series circuit. The cars are electric charges. The rate of traffic flow is current.

Capacitors change the rules a bit, adding a capacitor is like placing a rest stop next to the road, but there’s no way back out except by driving backwards up the offramp.

All of this stuff is required by conservation of mass & energy.

Anonymous 0 Comments

>If I = V/R, then in a series circuit, if the voltage is reducing after each component, then the current should also decrease right?

Not exactly. Voltage difference in electric potential and you need to measure it over the component. You can’t measure it to the ground or some other place. Ground what we define as 0 volts.

So the relevant voltage is the one you get if you measure with a voltmeter and put the probes on the two wires of the resistor. It is not the voltage between one of the wires and the ground.

I would say the water analogy have some problem. Some chase is better to compare to gravitational potential energy. Potential energy = mass * gravitational acceleration * height. Close to the surface of Earth, you can look at the elevation as the gravitational potential difference from what we define as zero elevation

If you drop a ball from the ceiling it will hit the floor at the same speed on the ground floor as it does on the first floor or in the basement. It does not matter where you are relative to the ground the house stands on it only matters how much the elevation changes

If you, on the other hand, drop the ball from the ceiling of the first floor to the floor of the ground floor it will gain 2x the kinetic energy because the gravitational potential difference is 2x higher. From the ceiling of the first floor to the basement is 3x the elevation and 3x the kinetic energy.

If you do it on the 3 top floors of a skyscraper the result is exactly the same. It do not matter you hare high above the ground, it is the elevation difference that matters

This assumes 0-floor thickness and that every room has the same ceiling height. It ignores air resistance. It also assumes the gravitation field is the same, it technically changes with elevation but it is a minuscule effect for short distances.

That you define 0 elevations as the ground does not matter is the difference that is relevant. I

Putting components in series is like splitting up the high something can drop.

There is a current reduction but it happening at the moment you add multiple components in series because the voltage drop is split over them, this means less voltage drop per component and lower current

If you have 20 V source and 2-ohm resistors. If you connect a single one you have 20V voltage drop over it and the current is 20/2= 10 amp.

If you connect two in series the voltage drop over each is 10V. The current is 10/2 = 5 amp.

You can also look at the whole circuit where the two resistors in series are added together and you have 4-ohm resistance. There is not a 20V voltage drop over 4 ohms and the current is 20/4=5 amp

You only have the voltage split up even if the restore has the same value. If you have a 3 and a 5-ohm resistor the voltage drop over them is not equal. The total resistance is 3+5= 8 ohms so the current is 20/8= 2.5 amp.

You can now calulate the voltage drop I=V/R => V=I * R This mean the voltage drop are 2.5 *3 =7.5V and 5 * 2.5 = 12.5V

The water pipe analog is not perfect

Anonymous 0 Comments

>If I = V/R, then in a series circuit, if the voltage is reducing after each component, then the current should also decrease right?

Not exactly. Voltage difference in electric potential and you need to measure it over the component. You can’t measure it to the ground or some other place. Ground what we define as 0 volts.

So the relevant voltage is the one you get if you measure with a voltmeter and put the probes on the two wires of the resistor. It is not the voltage between one of the wires and the ground.

I would say the water analogy have some problem. Some chase is better to compare to gravitational potential energy. Potential energy = mass * gravitational acceleration * height. Close to the surface of Earth, you can look at the elevation as the gravitational potential difference from what we define as zero elevation

If you drop a ball from the ceiling it will hit the floor at the same speed on the ground floor as it does on the first floor or in the basement. It does not matter where you are relative to the ground the house stands on it only matters how much the elevation changes

If you, on the other hand, drop the ball from the ceiling of the first floor to the floor of the ground floor it will gain 2x the kinetic energy because the gravitational potential difference is 2x higher. From the ceiling of the first floor to the basement is 3x the elevation and 3x the kinetic energy.

If you do it on the 3 top floors of a skyscraper the result is exactly the same. It do not matter you hare high above the ground, it is the elevation difference that matters

This assumes 0-floor thickness and that every room has the same ceiling height. It ignores air resistance. It also assumes the gravitation field is the same, it technically changes with elevation but it is a minuscule effect for short distances.

That you define 0 elevations as the ground does not matter is the difference that is relevant. I

Putting components in series is like splitting up the high something can drop.

There is a current reduction but it happening at the moment you add multiple components in series because the voltage drop is split over them, this means less voltage drop per component and lower current

If you have 20 V source and 2-ohm resistors. If you connect a single one you have 20V voltage drop over it and the current is 20/2= 10 amp.

If you connect two in series the voltage drop over each is 10V. The current is 10/2 = 5 amp.

You can also look at the whole circuit where the two resistors in series are added together and you have 4-ohm resistance. There is not a 20V voltage drop over 4 ohms and the current is 20/4=5 amp

You only have the voltage split up even if the restore has the same value. If you have a 3 and a 5-ohm resistor the voltage drop over them is not equal. The total resistance is 3+5= 8 ohms so the current is 20/8= 2.5 amp.

You can now calulate the voltage drop I=V/R => V=I * R This mean the voltage drop are 2.5 *3 =7.5V and 5 * 2.5 = 12.5V

The water pipe analog is not perfect

Anonymous 0 Comments

>If I = V/R, then in a series circuit, if the voltage is reducing after each component, then the current should also decrease right?

Not exactly. Voltage difference in electric potential and you need to measure it over the component. You can’t measure it to the ground or some other place. Ground what we define as 0 volts.

So the relevant voltage is the one you get if you measure with a voltmeter and put the probes on the two wires of the resistor. It is not the voltage between one of the wires and the ground.

I would say the water analogy have some problem. Some chase is better to compare to gravitational potential energy. Potential energy = mass * gravitational acceleration * height. Close to the surface of Earth, you can look at the elevation as the gravitational potential difference from what we define as zero elevation

If you drop a ball from the ceiling it will hit the floor at the same speed on the ground floor as it does on the first floor or in the basement. It does not matter where you are relative to the ground the house stands on it only matters how much the elevation changes

If you, on the other hand, drop the ball from the ceiling of the first floor to the floor of the ground floor it will gain 2x the kinetic energy because the gravitational potential difference is 2x higher. From the ceiling of the first floor to the basement is 3x the elevation and 3x the kinetic energy.

If you do it on the 3 top floors of a skyscraper the result is exactly the same. It do not matter you hare high above the ground, it is the elevation difference that matters

This assumes 0-floor thickness and that every room has the same ceiling height. It ignores air resistance. It also assumes the gravitation field is the same, it technically changes with elevation but it is a minuscule effect for short distances.

That you define 0 elevations as the ground does not matter is the difference that is relevant. I

Putting components in series is like splitting up the high something can drop.

There is a current reduction but it happening at the moment you add multiple components in series because the voltage drop is split over them, this means less voltage drop per component and lower current

If you have 20 V source and 2-ohm resistors. If you connect a single one you have 20V voltage drop over it and the current is 20/2= 10 amp.

If you connect two in series the voltage drop over each is 10V. The current is 10/2 = 5 amp.

You can also look at the whole circuit where the two resistors in series are added together and you have 4-ohm resistance. There is not a 20V voltage drop over 4 ohms and the current is 20/4=5 amp

You only have the voltage split up even if the restore has the same value. If you have a 3 and a 5-ohm resistor the voltage drop over them is not equal. The total resistance is 3+5= 8 ohms so the current is 20/8= 2.5 amp.

You can now calulate the voltage drop I=V/R => V=I * R This mean the voltage drop are 2.5 *3 =7.5V and 5 * 2.5 = 12.5V

The water pipe analog is not perfect

Anonymous 0 Comments

Current is how much charge passes through a cross-section of the wire per second.

In the case of electric wire it’s electrons, in batteries and such it’s charged particles, but it’s all the same.

So if you have a few components in series, the same amount of electrons passes through them every second, thus the same current passes through them

What gives rise to the voltage over each individual component is the current, not the other way around. All of these voltages add up to the voltage of the source, which is what gave rise to the current.

So source voltage causes current, current causes voltage over the load.