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Use the Proteus-style interactive schematic workbench below to add components with custom values, wire nodes, and inspect worked examples, or explore the focused analytical models.

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Add components with custom values, wire nodes, and run live nodal simulation.

01 · Current-Limited LED Indicator — VS + Switch + Series Resistor + Red LEDI_supply = 10.00 mA · P_total = 46.4 mW
VCC (5.0V)GND (0.0V)+−VS15 V10.00 mA · 5.00 VVCC (5.0V)N1 (5.0V)SW1CLOSED (ON)10.00 mA · 0.00 VN1 (5.0V)N2 (1.9V)R1330 Ω9.28 mA · 3.06 VN2 (1.9V)GND (0.0V)D1 (Red)Vf = 1.9 V9.28 mA · 1.94 VVCC: 5.00 VN1: 5.00 VN2: 1.94 VN3: 0.00 VN4: 0.00 VGND: 0.00 V
EDIT SELECTED COMPONENT
R1Fixed Resistor (R)
Live Component Status:

ΔV = 3.06 V · I = 9.28 mA · P = 28.4 mW

ADD COMPONENT TO CIRCUIT
DIFFERENTIAL MULTIMETER PROBE
V(N1) − V(GND) =5.00 V
Step-by-Step Design Calculation

01 · Current-Limited LED Indicator

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.

Given Parameters:

Supply Vs = 5.0 V, Red LED forward drop Vf = 1.9 V, Series resistor R1 = 330 Ω, Switch SW1 = Closed.

  1. 1. Apply Kirchhoff's Voltage Law (KVL) around the loop: V_R1 = Vs − Vf = 5.0 V − 1.9 V = 3.10 V.
  2. 2. Apply Ohm's Law to R1: I = V_R1 / R1 = 3.10 V / 330 Ω = 0.00939 A = 9.39 mA.
  3. 3. Compute resistor power dissipation: P_R1 = I² · R1 = (0.00939)² × 330 = 29.1 mW (well below ¼ W = 250 mW limit).

Engineering Result: Node voltages: V(VCC) = 5.00 V, V(N1) = 5.00 V, V(N2) = 1.90 V, V(GND) = 0.00 V.

ELECTRONIC COMPONENT ENCYCLOPEDIA & WIRING GUIDE

Fixed Resistor (R)

Linear passive component that opposes electric current and converts electrical energy into heat according to Ohm's Law.

V = I · R | P = I² · R = V² / R
1. How it works physically:

Current through a resistor is strictly proportional to the voltage difference across its two terminals: I = (Va − Vb) / R. It is non-polarized.

2. How to connect it in a circuit:

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.

3. Worked numerical 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).

4. Critical fault / mistake to avoid:

Forgetting to check power dissipation P = V² / R when using low-value resistors across 12 V or 24 V supplies.

Circuit workspaceLIVE MODEL
VCCAGND · 0 V5.0 VR1 · 330 ΩD1Vf ≈ 1.9 V
01 / LED + current-limiting resistorIdeal DC · fixed Vf model
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Predict before you measure.

What current do you expect through the LED?

Connect a virtual multimeterDMM
1.900 V

Black lead in COM. Ideal educational meter; input loading and fuse heating are not modeled.

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Bounded model: 0–5 V, 220–2200 Ω, Vf = 1.9 V. Brightness is illustrative. Hardware unverified.