Resistors drop voltage; they do not consume charge.
One of the deepest beginner misconceptions is thinking current gets 'used up' as it passes through a resistor. In reality, charge is conserved: every milliamp entering R1 leaves R1—what drops is electrical potential energy (voltage).
Dividing volts by kilohms directly yields milliamps. Keeping track of SI prefixes prevents 1000× calculation errors on the bench.
Three things to notice
- Conventional current direction. Schematics and equations track conventional current from higher potential (+) to lower potential (−).
- Voltage can exist without current. An open 5 V supply has 5 V across its terminals with 0 mA flowing.
- Check order of magnitude first. With a 5 V supply and resistors in the ratusan-to-thousands of ohms, currents live in the low milliamps.
Detailed Worked Engineering Example & Step-by-Step Solution
A 5 V DC source feeds a red LED (Vf = 1.9 V) through an SPST switch and a 330 Ω current-limiting resistor. Because the switch, resistor, and LED are in a single series loop, the exact same current flows through every component.
Supply Vs = 5.0 V, Red LED forward drop Vf = 1.9 V, Series resistor R1 = 330 Ω, Switch SW1 = Closed.
- 1. Apply Kirchhoff's Voltage Law (KVL) around the loop: V_R1 = Vs − Vf = 5.0 V − 1.9 V = 3.10 V.
- 2. Apply Ohm's Law to R1: I = V_R1 / R1 = 3.10 V / 330 Ω = 0.00939 A = 9.39 mA.
- 3. Compute resistor power dissipation: P_R1 = I² · R1 = (0.00939)² × 330 = 29.1 mW (well below ¼ W = 250 mW limit).
Conclusion: Node voltages: V(VCC) = 5.00 V, V(N1) = 5.00 V, V(N2) = 1.90 V, V(GND) = 0.00 V.
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
“Current is smaller after the resistor than before it.” — In a single loop, current is identical at every point simultaneously; the resistor limits the current of the entire loop.
“0.47 kΩ is 47 Ω.” — Multiply kilohms by 1000: 0.47 × 1000 = 470 Ω.
Learning outcomes, prerequisites & transfer question
- Distinguish voltage (across two nodes), current (through a cross-section), and resistance (opposition to flow).
- Convert accurately between A, mA, µA and Ω, kΩ, MΩ in circuit equations.
- Explain why current entering a series resistor equals the current leaving it.
Transfer check: If 6.60 mA enters the top lead of a 470 Ω resistor in a single series loop, how many milliamps exit the bottom lead, and what is the current in amperes?
Exactly 6.60 mA exits the bottom lead (charge conservation in a single loop), which equals 0.00660 A.