Is voltage constant in series? No. In a series circuit, current is constant, while voltage drops across each component in proportion to its resistance or impedance. The sum of these individual voltage drops equals the total source voltage, a principle defined by Kirchhoff’s Voltage Law (KVL). If you apply 12V to three identical 100Ω resistors in series, the voltage is not 12V at every point; it drops by 4V across each resistor, leaving 0V at the final ground node.
Understanding this distinction is the difference between a working prototype and a burned-out breadboard. Below, we break down the node-level physics, map out exactly what happens when components fail, and walk through a real-world 12V design.
Series Topology and Node Analysis
A series topology forces electrons down a single, continuous path. Because there are no alternate branches, the exact same number of electrons per second (current, measured in Amperes) must flow through every component. However, as current pushes through resistive elements, electrical potential energy is converted into heat or light. This loss of potential is the voltage drop.
To analyze this, we label the connection points as nodes. Consider a simple circuit with a 12V DC source and three resistors (R1, R2, R3):
- Node A: Positive terminal of the 12V source (12.0V potential).
- Node B: Junction between R1 and R2.
- Node C: Junction between R2 and R3.
- Node D: Negative terminal / Ground (0.0V potential).
If R1, R2, and R3 are all 1kΩ, the total resistance is 3kΩ. The constant circuit current is $I = V / R = 12V / 3000Ω = 4mA$. The voltage at Node B will be 8.0V (having dropped 4V across R1). Node C will be 4.0V (dropping another 4V across R2). Node D is 0V. As detailed in the All About Circuits DC textbook, KVL dictates that the algebraic sum of all drops in the loop must equal the source voltage.
You use series topologies when you need current limiting or voltage division. Wiring high-voltage LED strings in series allows a single current-limiting resistor to regulate the entire chain, saving power and component count. Parallel circuits, conversely, maintain constant voltage across branches but require individual current management for each leg.
Behavior Matrix and Failure Extremes
The most common mistake hobbyists make is assuming a series circuit will gracefully handle component variations. It will not. Because current is shared, a change in one component alters the voltage distribution for all components.
| Component Change | Total Resistance ($R_T$) | Circuit Current ($I$) | Voltage Drop Across Other Components |
|---|---|---|---|
| Increase R1 | Increases | Decreases | Decreases (R1 hogs more voltage) |
| Decrease R1 | Decreases | Increases | Increases (R1 drops less voltage) |
| Open R1 (Break) | Infinite ($\infty$) | Zero (0A) | Zero (0V across others; full source voltage appears across the open break) |
| Short R1 (Bypass) | Decreases heavily | Spikes | Spikes (Remaining components absorb full source voltage, usually leading to thermal failure) |
The Open-Circuit Extreme: If a wire breaks or a fuse blows at R1, current stops immediately. If you probe Node B with a digital multimeter (DMM) referenced to ground, it will read 12V. The DMM's high input impedance (typically 10MΩ) completes the circuit just enough to measure the source potential, but no real current flows.
The Short-Circuit Extreme: If R1 fails short (0Ω), Node B effectively becomes Node A. The remaining components (R2 and R3) now divide the full 12V instead of their designed 8V. If they are rated for 1/4W, this voltage spike will likely exceed their power dissipation limits ($P = V^2/R$), causing a cascading thermal failure.
Design Walkthrough: 12V LED Current-Limiting String
Let’s apply this theory to a practical build. We want to wire three standard 5mm red LEDs in series from a 12V DC bench supply. According to the SparkFun series/parallel tutorial, LEDs are current-driven devices; they will draw as much current as the source can provide until they burn out, making a series current-limiting resistor mandatory.
1. Define Component Values:
- Source Voltage ($V_S$): 12.0V DC
- LED Forward Voltage ($V_f$): 2.1V typical per red LED
- LED Target Current ($I_f$): 20mA (0.020A)
2. Calculate the Voltage Drop Required for the Resistor:
The three LEDs in series will consume $3 \times 2.1V = 6.3V$. The remaining voltage must be dropped across the current-limiting resistor.
$V_R = 12.0V - 6.3V = 5.7V$
3. Calculate Resistance (Ohm's Law):
$R = V_R / I_f = 5.7V / 0.020A = 285Ω$
4. Select Real-World Components:
285Ω is not a standard value. Using the E12 resistor series, the nearest standard value is 330Ω. Let's recalculate the actual current with 330Ω:
$I_{actual} = 5.7V / 330Ω = 17.2mA$
This is perfectly safe and will yield a slightly dimmer, longer-lasting LED.
5. Verify Power Rating:
$P = I^2 \times R = (0.0172A)^2 \times 330Ω = 0.097W$
A standard 1/4W (0.25W) carbon film resistor is more than adequate, operating at less than 40% of its thermal limit.
Step-by-Step Breadboard Verification
Do not trust your math until you verify it on the bench. Here is how to test the node voltages using an auto-ranging DMM.
- Insert the Resistor: Place the 330Ω resistor across the breadboard's center trench. Connect one leg to the positive (red) power rail via a jumper wire.
- Chain the LEDs: Insert LED1 so its anode (long leg) connects to the other leg of the resistor. Connect LED1's cathode (short leg) to LED2's anode in the same column. Repeat for LED3.
- Complete the Circuit: Connect LED3's cathode to the negative (blue) ground rail. Ensure the bench supply is OFF.
- Power and Measure Source: Turn on the 12V supply. Set your DMM to DC Volts. Place the black probe on the ground rail and the red probe on the positive rail. Verify it reads ~12.0V (Node A).
- Measure Node Drops: Keep the black probe on ground. Move the red probe to the junction between the resistor and LED1. You should read ~5.7V. Move to the next LED junction: ~3.6V. Move to the final cathode: ~1.5V to 2.0V (accounting for minor breadboard contact resistance and LED $V_f$ tolerances).
- Measure Current: Turn off the supply. Break the circuit at the ground rail, insert the DMM in series (set to mA), and power on. Verify the reading is ~17.2mA.
Frequently Asked Questions
Is voltage the same across all components in a series circuit?
No. Voltage is only identical across series components if their resistance or impedance values are exactly matched. If you place a 100Ω resistor and a 1kΩ resistor in series across a 12V source, the 1kΩ resistor will drop roughly 10.9V, while the 100Ω resistor will drop only 1.1V. The voltage distributes proportionally to the resistance ($V_x = V_{total} \times (R_x / R_{total})$).
Why does voltage drop in a series circuit but current stays the same?
Think of current as the flow rate of water in a single pipe, and voltage as the water pressure. If the pipe has no leaks or branches (series), the exact same volume of water passes every point per second (constant current). However, as the water pushes through narrow restrictions (resistors), it loses pressure (voltage drop) due to friction. The pressure is expended to push the flow through the restriction, but the flow rate itself remains continuous.
Does adding more resistors in series increase or decrease the total voltage?
Adding resistors in series does not change the source voltage; your power supply still outputs exactly what it is set to (e.g., 12V). However, adding more resistors increases the total circuit resistance, which lowers the overall current. Consequently, the voltage drop across each individual resistor will decrease, but the sum of all the drops will still perfectly equal the source voltage.
What happens to the voltage if one bulb blows in a series string?
When a bulb blows, its filament breaks, creating an open circuit. The current instantly drops to zero. Because there is no current flowing through the remaining good bulbs, their voltage drops become 0V ($V = I \times R$, and $I = 0$). By KVL, the entire source voltage now appears across the broken bulb's gap. This is why, in old-school series-wired Christmas lights, a single dead bulb kills the whole string, and measuring across the dead socket will show the full 120V AC line potential.
