Observe the circuit without becoming the load.
A voltmeter connects in parallel across two nodes. Because its internal resistance is around 10 MΩ, almost zero current diverts into the meter—unless you leave the red lead plugged into the current port!
If the red probe sits in the mA/A jack while you probe across a voltage source, the meter's low-resistance current shunt creates an immediate short circuit.
Three things to notice
- Voltage is always across two points. You do not break the circuit to measure voltage; touch red to the target node and black to your reference.
- Sign tells you polarity. Swapping red and black probes flips the sign from +3.10 V to −3.10 V without harming a digital multimeter.
- OL or floating readings. An unconnected node has no defined potential path to reference.
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
“Voltage flows through a component.” — Current flows through components; voltage is the potential difference across two terminals.
“Turning the dial to DC V makes the probes safe regardless of jack.” — The physical banana jack determines whether the red lead connects to the 10 MΩ divider or the low-resistance current fuse.
Learning outcomes, prerequisites & transfer question
- Configure a DMM with the black lead in COM, red lead in V/Ω, and dial set to DC Volts.
- Measure node voltages and component voltage drops in parallel across two points.
- Explain why a voltmeter has very high input resistance (typically 10 MΩ).
Transfer check: You measure 5.00 V from VCC to GND, and 5.00 V from node A to GND in a series R1 + LED circuit. What is the voltage across R1, and what does that tell you about the current?
The voltage across R1 is V_VCC − V_A = 5.00 − 5.00 = 0.00 V. By Ohm's law (I = V/R), zero voltage across R1 proves zero current is flowing (e.g., LED reversed or open path to GND).