A voltage divider is not a power regulator.
Two equal 1 kΩ resistors divide 5.00 V neatly into 2.50 V—as long as nothing draws current from the midpoint. Connect a 1 kΩ load across the bottom resistor and the lower branch drops to 500 Ω, pulling V_out down to 1.67 V!
To keep loading droop under 1%, the load resistance must be at least 100× larger than the divider's Thevenin resistance (R1 ∥ R2).
Three things to notice
- Thevenin viewpoint. A 5 V, 1 kΩ / 1 kΩ divider behaves just like an ideal 2.5 V source in series with R_th = 1 kΩ ∥ 1 kΩ = 500 Ω.
- Even your DMM is a load. On a 1 kΩ divider, a 10 MΩ DMM causes negligible error; on a 10 MΩ divider, the same DMM pulls the reading down by 33%!
- Sensors as variable dividers. Thermistors (NTC) and photoresistors (LDR) pair with a fixed resistor to convert physical changes into node voltages.
Detailed Worked Engineering Example & Step-by-Step Solution
Two 1 kΩ resistors divide 10 V into 5.00 V unloaded. Closing switch SW_L connects a 1 kΩ load in parallel with R2, collapsing the equivalent lower resistance to 500 Ω and pulling the output down to 3.33 V!
Vs = 10.0 V, R1 = 1000 Ω, R2 = 1000 Ω, Load R_L = 1000 Ω.
- 1. Unloaded (SW_L Open): V_out = Vs · R2 / (R1 + R2) = 10 × 1000 / 2000 = 5.00 V.
- 2. Loaded (SW_L Closed): R_eq = (R2 × R_L) / (R2 + R_L) = (1000 × 1000) / 2000 = 500 Ω.
- 3. New Output Voltage: V_out(loaded) = 10 × 500 / (1000 + 500) = 3.33 V (a 33.3% voltage droop!).
Conclusion: Rule of thumb: For < 10% droop, load resistance R_L must be at least 10× larger than R2.
Electronic Components Reference: How Each Part Works & Connects
An ideal DC source holds V(+) − V(−) = Vs regardless of load current. Real bench supplies have a current limit (CC mode) and small internal resistance.
How to connect: Connect nodeA (+) to your main supply rail (VCC) and nodeB (−) to the reference ground node (GND = 0 V). Never wire two different voltage sources directly in parallel across the same two nodes.
Worked Example: Example: A 9 V source connected across a 450 Ω load resistor drives I = 9 V / 450 Ω = 20 mA and delivers P = V · I = 9 V × 0.02 A = 180 mW.
Current through a resistor is strictly proportional to the voltage difference across its two terminals: I = (Va − Vb) / R. It is non-polarized.
How to connect: Place in series with loads (such as LEDs or Zener diodes) to limit current, or in pairs across a supply to form a voltage divider.
Worked Example: Example: To limit a 2.0 V LED to 10 mA from a 5 V supply: R = (5 − 2.0) / 0.01 A = 300 Ω (choose standard 330 Ω). Power: P = (0.01)² × 330 = 33 mW (safe for a 250 mW ¼ W resistor).
When wired as a rheostat (variable resistance between nodeA and nodeB), R_eff = R_total × (wiper / 100).
How to connect: Connect in series with a fixed minimum stopper resistor (e.g., 100 Ω) so turning the knob to 0% never shorts the supply.
Worked Example: Example: A 1 kΩ potentiometer set to α = 50% gives R_eff = 500 Ω. In series with a 220 Ω protection resistor, total resistance is 720 Ω.
An open switch drops the full available loop voltage across its open contacts while passing 0 mA. A closed switch passes loop current with negligible voltage drop (≈ 0 V).
How to connect: Place in series on the high-side (between VCC and load) to control power delivery to the circuit branch.
Worked Example: Example: In a 5 V LED circuit, opening SW1 makes I = 0 mA; measuring across SW1 with a voltmeter reads 5.00 V because the anode side sits at 5 V and the load side drops to 0 V.
Unlike a resistor, an LED has an exponential I–V curve. Once forward voltage reaches Vf, its dynamic resistance drops to a few ohms and it cannot limit its own current.
How to connect: Connect Anode (nodeA, triangle base) toward positive potential and Cathode (nodeB, bar) toward GND, ALWAYS in series with a current-limiting resistor.
Worked Example: Example: Red LED (Vf = 1.9 V) with R = 330 Ω on a 5.0 V supply: V_R = 5.0 − 1.9 = 3.1 V, so I = 3.1 V / 330 Ω = 9.39 mA (bright, safe operation below 20 mA max).
When V(Anode) − V(Cathode) ≥ 0.7 V, the depletion barrier collapses and current flows. When reverse-biased, only negligible leakage current flows.
How to connect: Place in series with the positive supply input for reverse-polarity protection, or across inductive loads (cathode to +V) as a flyback protection diode.
Worked Example: Example: A 5.0 V supply passes through a series 1N4007 protection diode (Vf = 0.7 V) to a 430 Ω resistor: downstream node voltage is 5.0 − 0.7 = 4.3 V, and current is 4.3 / 430 = 10.0 mA.
Common misconceptions & corrections
“A 1 kΩ / 1 kΩ divider at 5 V can power a 2.5 V motor or LED at a steady 2.5 V.” — Any low-resistance load pulls the divider voltage down sharply because current must pass through R1.
“V_out is always V_in × R1 / (R1 + R2).” — When V_out is measured from the midpoint to GND, the numerator is the bottom resistor R2 (between OUT and GND), not top resistor R1.
Learning outcomes, prerequisites & transfer question
- Calculate the unloaded output of a two-resistor voltage divider: V_out = V_in × R2 / (R1 + R2).
- Calculate loaded output voltage by combining R2 in parallel with R_load.
- Explain why a resistive voltage divider is suitable for high-impedance sensor/ADC inputs, not heavy power loads.
Transfer check: You have a 5.0 V supply and want an unloaded divider output of 1.0 V using a bottom resistor R2 = 1.0 kΩ. What value of top resistor R1 is required?
Since V_out = 1.0 V across R2 = 1.0 kΩ, the current is 1.0 mA. R1 must drop the remaining 4.0 V at 1.0 mA, so R1 = 4.0 V / 1.0 mA = 4.0 kΩ.