Apply V = I × R to the right two nodes.
Ohm's law is simple, yet learners frequently plug the entire supply voltage into I = V / R when an LED or second component shares the loop. The V in V = I × R is strictly the voltage difference across that resistor.
Double the voltage across a fixed resistor (within its power rating) and the current doubles. Across an LED, a tiny voltage increase above ~1.9 V causes a steep current surge.
Three things to notice
- Steeper I–V slope means lower resistance. On an I-versus-V plot, a 330 Ω resistor has a steeper slope than a 1 kΩ resistor because it passes more current per volt.
- Temperature limits linearity. Even a metal-film resistor changes resistance slightly if overheated; an incandescent lamp filament changes dramatically.
- Verify by two measurements. Measure R with power off, then measure V_R with power on to cross-check predicted current.
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
“Every component obeys R = V / I with a constant R.” — LEDs, diodes, and transistors are non-linear devices whose voltage-current relationship is exponential, not a fixed resistance.
“In V = I × R, V is always the power supply voltage.” — V is the voltage drop across the specific resistor being analyzed.
Learning outcomes, prerequisites & transfer question
- Solve for V, I, or R across an ohmic resistor using V_R = I × R.
- Interpret the slope of an I–V graph as conductance (1/R).
- Explain why Ohm's law applies to the resistor's voltage drop, not directly to an LED.
Transfer check: A resistor passes 4.0 mA when the voltage across its leads is 2.0 V, and 8.0 mA when the voltage across its leads is 4.0 V. What is its resistance, and what current flows at 5.0 V?
R = 2.0 V / 0.004 A = 500 Ω. Because it is linear, at 5.0 V the current is 5.0 V / 500 Ω = 10.0 mA.