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I know that this may seem like a simple question but I want to know whether the currents across the resistors of different values are equivalent.

For example in the circuit depicted below, if I were to measure the current just after R1 and just after R2, would the currents be the same?

There is no other path for the electrons to travel but the electrons hit atoms in the resistor material so possibly they are slowed down.

circuit with two resistors

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  • \$\begingroup\$ Comments are not for extended discussion; this conversation has been moved to chat. \$\endgroup\$
    – Voltage Spike
    Feb 27 at 23:28
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    \$\begingroup\$ I'm pretty sure this user is trolling on this post \$\endgroup\$
    – Voltage Spike
    Feb 27 at 23:30
  • \$\begingroup\$ It's probably worth noting that there is no point to having a voltage divider with no load, and the design requirement is that the current be negligible compared to the voltage divider's constant current. When you decide you need a voltage divider, the minimum amount of power you waste in the divider is determined by the size of the load, so say it's a comparator input, you can look up the input current of the comparator and make the voltage divider's current 100x or 1000x that amount to ensure the divider stays accurate over that load range. \$\endgroup\$
    – K H
    Feb 28 at 2:16
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There's no other path for the electrons to travel

This is the correct answer. The same current is going through both resistors.

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    \$\begingroup\$ At least as long as V_out is not used, i.e. nothing is connected there in parallel to R2. \$\endgroup\$ Feb 27 at 18:41
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There is no other path for the electrons to travel

Therefore the current must be the same for every element in a loop -- whether they're resistors, batteries, diodes, or butterflies.

If the current weren't the same in every element, the electrons would pile up somewhere, and since electrons repel each other, they'd find a way to get through.

but the electrons hit atoms in the resistor material so possibly they are slowed down.

It's more complicated than that, but -- yes! It's just that because the current must be equal in every device in the loop, a highly resistive device slows the electrons down for everything in the loop (possibly to the advantage of the butterfly, depending on the source voltage).

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Assuming there is nothing connected to the Vout terminals, then yes, the current is the same in both resistors. Also, BTW, electrons do not "slow down".

Intuitively, think of it this way. The mobile electrons have to go somewhere. At two different points on a highway with no exits in-between, all of the cars going past point A have to go past point B. In a simple series circuit with ((any)) number of components of any type (resistors, capacitors, inductors, diodes, whatever), 100% of the current goes through 100% of the components 100% of the time. The nature of the components has a direct effect on what that current is, but all of the energy is "trapped' in the loop.

For a string of resistors, the current is determined by the total resistance of the string. Inside the string, the voltage across each individual resistor is directly proportional to its resistance, per Ohm's Law. As an exercise, draw up a string of four resistors in series with the values of 1, 2, 3, and 4 ohms, connected to a 10 V battery. First, calculate the current through the string, then use that current to calculate the voltages across each resistor, then add up those individual voltages and compare that number to the battery voltage.

A similar thing is true with parallel circuits. With 2 or more resistors in parallel, 100% of the voltage appears across 100% of the components 100% of the time. This time, again using Ohm's Law, the current through each resistor is inversely proportional to the value of the resistor.

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