Correct ohms can still burn if the wattage is too small.
Every milliwatt dissipated in a resistor turns directly into heat. A standard through-hole resistor is rated for 0.25 W (250 mW) in free air, but good engineering derates continuous dissipation to half of the maximum rating.
Doubling the current through a resistor quadruples its heat dissipation. Always check both ohms and milliwatts before powering up.
Three things to notice
- Rating is a ceiling, not actual power. A 0.25 W resistor in our LED circuit dissipates only ~29 mW, not 250 mW.
- Why derate by 50%? At 100% rated power, a resistor's surface can exceed 100 °C—hot enough to burn fingers and drift in value.
- Tolerance shifts current. A 330 Ω ±5% resistor can be as low as 313.5 Ω, raising LED current from 9.39 mA to 9.89 mA under the fixed-Vf model.
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 1/4 W resistor always consumes 0.25 W.” — Actual dissipation is set by V_R² / R in the circuit; 0.25 W is the maximum mechanical/thermal limit.
“Doubling supply voltage doubles resistor power.” — Because P = V² / R, doubling the voltage across a fixed resistor multiplies its power by four.
Learning outcomes, prerequisites & transfer question
- Compute resistor power dissipation using P = V_R × I = I² × R = V_R² / R.
- Apply 50% power derating (e.g., ≤ 125 mW continuous on a 0.25 W resistor).
- Calculate worst-case current bounds from ±5% or ±1% resistor tolerance.
Transfer check: You need to place a 100 Ω resistor directly across a 5.0 V rail. How much power will it dissipate, and is a 0.25 W resistor acceptable under a 50% derating rule?
P = 5² / 100 = 0.25 W (250 mW). Under a 50% derating rule, a 0.25 W resistor should not exceed 125 mW, so you must select at least a 0.5 W resistor.