Why does voltage drop in a series circuit but current stays the same

circuit analysiscurrentvoltage

my basic understanding of electricity is as follows:

Voltage is the difference between the electric field strength of 2 points and how much work can be done by a charged particle as it is moved by those fields between those 2 points.

Current is the flow of charged particles and the consequence of voltage. There can be no current without voltage although there can be voltage without current.

Resistance is a measure of how much a material inhibits current.

The relationship between these three is expressed in an equation:

V = I x R

The problem I have is that when I apply this understanding to a series circuit, I struggle to wrap my head around how current and resistance can remain constant throughout the circuit but voltage varies? Shouldn't the current drop with the voltage? There are other people who have asked similar question but I could not really understand the answers so please keep it simple for me. Thank you.

Best Answer

I struggle to wrap my head around how current and resistance can remain constant throughout the [series] circuit but voltage varies? Shouldn't the current drop with the voltage?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Two 2 kΩ loads on a 10 V supply.

  1. First thing: for a resistor, \$ I_{IN} = I_{OUT} \$. Otherwise mobile charges would pile up somewhere inside the component and this doesn't happen.
  2. As a result of 1, current in a series circuit is the same throughout. In Figure 1 \$ I_{R1} = I_{R2} \$ and \$ I_{R3} = I_{R4} = I_{R5} \$.
  3. Voltages is measured between two points. It is common to measure with respect to an arbitrary ground (battery negative in Figure 1) but can also be measured between any two points. VM3 shows the voltage at the top of R1 with respect to ground while VM1 shows the voltage between the terminals of R1.
  4. The circuit consisting of R3, 4 and 5 has the same overall resistance but notice the way the voltage is divided in proportion to the value of the resistors.

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Figure 2. See my answer to Intuitive interpretation of negative voltage for more on voltage references. Image original by @Transistor.

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