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Lesson DC Circuits · Series and dividers

Series voltage drop

In series, the source voltage does not fall entirely at one point—it splits. Each resistor takes a portion (E = I imes R). The drops add back to the applied voltage. Where there are more ohms, there is more voltage drop at the same current.

1

Why voltage drop appears

When current flows through a resistance, a potential difference is needed to push that current. That difference is the voltage drop across the element: \[ E = I imes R \] In series, \(I\) is the same everywhere. Therefore whoever has more R gets more E.

2

The sum rule

On a closed resistive DC loop: \[ E_T = E_1 + E_2 + E_3 + \ldots \] This is part of Kirchhoff's voltage law, developed fully in Unit 9. For now: drops must account for the full applied voltage.

Example: \(I_T = 0.020\,\mathrm{A}\), \(R_1 = 1000\,\Omega\), \(R_2 = 1800\,\Omega\), \(R_3 = 2000\,\Omega\), \(R_4 = 1200\,\Omega\). \[ E_1 = 20\,\mathrm{V},\; E_2 = 36\,\mathrm{V},\; E_3 = 40\,\mathrm{V},\; E_4 = 24\,\mathrm{V} \] \[ E_T = 20 + 36 + 40 + 24 = 120\,\mathrm{V} \] If your drops do not sum to \(E_T\) (within meter tolerance), you measured wrong or missed a branch.

3

Proportional drops when values differ

Four equal resistors on 24 V: each drop is 6 V (24/4).

When values differ, each drop is proportional to its resistance at the same current. A 14 Ω resistor carries a larger share of 24 V than an 8 Ω resistor in the same loop.

Sketch polarity: mark "+" on the end where conventional current enters each resistor. That keeps diagram readings aligned with meter signs.

4

How to measure drops in the field

With the circuit energized (proper PPE and CAT rating):

  1. Pick a return reference (common negative, control ground, etc.).
  2. Measure voltage across each component (tip to tip), or node to node.
  3. Record polarity: in conventional DC, the end toward the positive source side is at higher potential.

You may also measure from the return upward and subtract between nodes. What matters is knowing what you are measuring.

5

Polarity of drops in one loop

In a single-source series loop, resistor drops oppose the source when you walk the loop. Sum them algebraically and you get zero—Kirchhoff's idea.

For a new technician: mark "+" where current enters each R and "−" where it leaves. That prevents sign confusion when comparing to the print.

6

Wire and connections drop voltage too

A loose connection or long thin wire is extra series resistance. Same \(I\) → unwanted drop → the load sees less voltage than you think.

Typical symptom: 24 V at the supply; 18 V at the coil; the rest lost in cables and contacts. Not magic: \(E = I imes R_{ ext{parasitic}}\).

This is why we care about voltage drop on conductors in a later course—not only on labeled resistors.

7

Calculation order that works

Known \(E_T\) and all R:

  1. \(R_T = \sum R\)
  2. \(I_T = E_T / R_T\)
  3. \(E_n = I_T imes R_n\) for each part
  4. Verify \(\sum E_n = E_T\)

Known \(I_T\) and all R (current limited supply):

  • Skip step 2; go straight to drops.

Known one drop and one R:

  • \(I = E / R\), then propagate through the series chain.
8

Using drops to find opens and shorts

Open in series: current → 0; you may read full \(E_T\) across the open; other drops → 0.

Short across one R: that branch drops → 0; its share redistributes—often higher current and larger drops elsewhere.

Walk the chain with a voltmeter: the element eating almost all the voltage is often the open (or the load when everything else is good).

9

Field case

Situation. A 120 V DC control circuit feeds a relay coil (500 Ω) through 30 ft of #18 wire (total loop resistance including connections measured as 8 Ω). The coil chatters.

How to think with this lesson.

  • Total R ≈ 508 Ω → \(I ≈ 120/508 ≈ 0.236\,\mathrm{A}\).
  • Wire drop: \(E_{ ext{wire}} = 0.236 × 8 ≈ 1.9\,\mathrm{V}\).
  • Coil sees ≈ 118 V—not full 120 V.
  • Under load, supply may sag further; coil pull-in margin shrinks.

Conclusion: series drops are not "lost" mysteriously—they are \(I × R\) somewhere. Find where voltage is missing.

In the field

Symptom

Load weak or off; source reads normal unloaded

Where to look

Series path: wire, fuse, contacts, coil, return

Likely causes

  1. Excessive I×R in wire/connections
  2. open in chain
  3. partial open (corroded contact)

What to measure

  1. V across each series element under load
  2. compare sum to ET

What not to do

  • Measure only at the supply and assume the load gets the same

Checklist

  • I use E = I × R for each series element
  • I verify drops sum to applied voltage
  • I measure across components, not only to ground (unless intentional)
  • I mark polarity on my sketch
  • I include wire and connections as series R
  • I know an open often shows full ET across it

Common mistakes

Symptom Typical cause Action
Drops sum to 90 V on a 120 V loop Meter on wrong scale; missed element; parallel path Measure every series element; re-trace diagram
Zero V across coil, full V at supply Open somewhere in series; measuring line to line only Walk the chain under load
Ignore small wire drop Assumed wire is ideal Calculate or measure I×R on conductors
Polarity confusion on print Mixed reference points One reference; mark + on each R