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I have some slight confusion about voltage drop across electronic components.

If the current before and after the resistor is the same, why is there a voltage drop across the input and output terminals of the resistor?

Voltage drop is the joules of energy lost when moving 1 coulomb of charge across 1 ohm. So its essentially a force which acts on electrons at a point.

If the current is the same before and after the resistor, why is there a difference of electromotive force?

The only other thought i have may be that the drop is compared to when the resistor is not there

Can anyone clarify these concepts please?>

enter image description here

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    \$\begingroup\$ What does OHm’s Law tell you? \$\endgroup\$ May 25 at 13:04
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    \$\begingroup\$ Does this answer your question? What exactly is voltage? \$\endgroup\$
    – Mitu Raj
    May 25 at 13:05
  • \$\begingroup\$ Voltage is a measure of energy per unit charge [J/C]. When a unit of charge passes through a resistor and heats it up, it has lost some of its energy. \$\endgroup\$
    – vir
    May 25 at 13:13
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    \$\begingroup\$ Maybe the answers to this related question will help: electronics.stackexchange.com/questions/566529/… \$\endgroup\$
    – ErikR
    May 25 at 13:16
  • \$\begingroup\$ V= I * R........ \$\endgroup\$ May 25 at 13:33
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Electric Potential Difference OR Voltage:

enter image description here

• Consider the task of moving a positive test charge within a uniform electric field from location A to location B as shown in the diagram.

• In moving the charge against the electric field from location A to location B, work will have to be done on the charge by an external force.

• The work done on the charge changes its potential energy to a higher value; and the amount of work that is done is equal to the change in the potential energy.

• As a result of this change in potential energy, there is also a difference in electric potential between locations A and B.

• This difference in electric potential is represented by the symbol ΔV and is formally referred to as the electric potential difference.

• By definition, the electric potential difference is the difference in electric potential (V) between the final and the initial location when work is done upon a charge to change its potential energy.

• Because electric potential difference is expressed in units of volts, it is sometimes referred to as the voltage.

Current:

• Current has to do with the number of coulombs of charges that pass a point in the circuit per unit of time (not with how much of potential energy that charge carries).

Answer:

As the charge moves through the circuit it loses its electric potential hence voltage drops. The charge is simply the medium which moves the energy from location to location.

While the energy possessed by the charge may be used up (or say that the electric energy is transformed to other forms of energy), the charge carriers themselves do not disintegrate, disappear or otherwise become removed from the circuit.

That's why voltage drops but current is same (charge carriers that pass a point does not change only their energy becomes used up)

Images used and explanations from physicsclassroom.com website

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  • \$\begingroup\$ Exactly how is the energy possessed by the charged used up? It collides with atoms with high kinetic energy. This kinetic energy is produced by an electromotive force which is produced by a higher concentration of charged particles at the anode terminals a battery supply. Still, if the current before and after the resistor remains unchanged, how can electric potential energy be lost? The potential energy is what causes a current or movement fo charged particles \$\endgroup\$
    – Malemna
    May 25 at 14:03
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    \$\begingroup\$ Do not get confused between charges and energy possessed by these charges. Even if the energy possessed by the charge is used up by the load (resistance or collisions with other atoms in the wire) still the charge carriers itself are present. Suppose if you have an object at a height h on table (It has potential energy due to virtue of height and no kinetic energy as its at rest). When you drop it to ground, potential energy changes to kinetic energy and as it drops to ground, it transfers it energy to the ground and its energy gets used up. \$\endgroup\$
    – user284706
    May 25 at 14:20
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    \$\begingroup\$ [Continuation] Only its energy is getting used up, Object itself is still present on the ground. You can apply this analogy to charges too. Only energy possessed by the charges is getting used up due to collisions (hence voltage drops). But charges itself are present (current is numbers of coulombs of charges that pass a point in the circuit). Only energy possessed by the charges is getting used up. \$\endgroup\$
    – user284706
    May 25 at 14:21
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    \$\begingroup\$ [Continuation] Inside the battery, charge carriers gain energy (through electrodes and electrolyte action). When battery is connected to a circuit, this energy gained in battery is lost (transformed to other forms of energy heat, light etc) by the charges as they travel through wires, resistors (bulbs)etc. Only energy is used up (transformed to other forms] Charges itself are present in the circuit. \$\endgroup\$
    – user284706
    May 25 at 14:28
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    \$\begingroup\$ @Malemna - Hi, Polite technical criticism of an answer (or question) is allowed, but please be careful not to cross the line into an "ad hominem" attack of a person. Therefore one of your comments has been deleted. I recommend you re-read the Code of Conduct. Thanks for your help with this. \$\endgroup\$
    – SamGibson
    May 26 at 12:19

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