Ohm's Law Circuit Solver: V, I, R + Series/Parallel
Solve voltage, current, resistance, and power. Calculate series/parallel equivalents, voltage dividers, and current dividers with interactive visualizations.
Georg Simon Ohm published Die galvanische Kette, mathematisch bearbeitet in 1827. The result was V = IR for what we now call ohmic conductors. The relationship had been suspected by Cavendish forty years earlier (he never published), but Ohm gave the first systematic measurements and the linear law that bears his name. Modern physics treats Ohm's law as the small-signal limit of a more general I-V curve. Anything where current is linear in applied voltage qualifies as ohmic. Most metallic resistors and doped semiconductors at low fields are. Diodes, transistors, plasma columns, and gas discharges aren't, which is why those devices are useful for switching and amplification.
Ohm's Law Formulas at a Glance
| To Find | Formula | You Need |
|---|---|---|
| Voltage (V) | V = I × R | Current and resistance |
| Current (I) | I = V / R | Voltage and resistance |
| Resistance (R) | R = V / I | Voltage and current |
| Power (P) | P = V × I | Voltage and current |
| Power (P) | P = I² × R | Current and resistance |
| Power (P) | P = V² / R | Voltage and resistance |
Defining Polarity and Current Direction Before You Solve
Ohm's law is a scalar equation but you're applying it to a directed quantity. Before you write V = IR, pick a reference direction for current and a reference polarity for voltage. The two have to agree, or your sign will flip and you won't notice until the answer doesn't match the expected battery voltage.
The standard rule is the passive sign convention. Mark the resistor with a + on the terminal current enters and a − on the terminal it leaves. Then V = IR comes out positive for power being absorbed. If you flip the current arrow, V comes out negative. Real instruments don't care about your convention, so a probe reading −3.4 V across a 1 kΩ resistor with current entering the side you marked − just means current is flowing the other way.
Conventional current flows from + to − through a resistor in the external circuit. Electron flow is the reverse. Both conventions exist in textbooks and you'll see arrows drawn either way. Pick one and stick with it for the whole problem. Mixing them once is enough to get every sign wrong.
V = I × R (Voltage = Current × Resistance)
I = V / R (Current = Voltage ÷ Resistance)
R = V / I (Resistance = Voltage ÷ Current)
Units: voltage in volts (V), current in amperes (A), resistance in ohms (Ω). The most common error is mixing mA with A. Divide mA by 1000 before plugging in. 20 mA is 0.020 A, not 20.
Quick check: If your answer seems off by 1000×, you probably forgot a unit conversion. A 9 V battery through 1 kΩ gives 9 mA, not 9 A.
Series vs. Parallel: A Mental Model for Voltage and Current Sharing
The cleanest way to think about series and parallel: in series, current is shared (everyone gets the same I, voltage divides). In parallel, voltage is shared (everyone gets the same V, current divides). Once you internalize that, you don't need to memorize the combination formulas. They fall out of Kirchhoff's laws.
Series
Same current through all. Voltages add.
R_total = R1 + R2 + R3 + ...
10 Ω + 20 Ω + 30 Ω = 60 Ω
Parallel
Same voltage across all. Currents add.
1/R_total = 1/R1 + 1/R2 + 1/R3 + ...
Two 100 Ω in parallel = 50 Ω (not 200 Ω).
For just two resistors in parallel, there's a shortcut: R_total = (R1 × R2) / (R1 + R2). Two equal resistors in parallel give half of one. So 100 Ω parallel with 100 Ω is 50 Ω.
A voltage divider is the canonical series example. Two resistors R1 and R2 across a supply, output taken at the junction: V_out = V_in × R2 / (R1 + R2). Creating 3.3 V from 5 V: target ratio 3.3/5 = 0.66, so pick R2 = 10 kΩ and R1 ≈ 5.1 kΩ. The output is 3.31 V.
Loading warning: Connecting a load across R2 creates a parallel combination, lowering effective R2 and dropping V_out. Keep divider resistance much lower than load impedance (10× rule), or buffer with a TL072 op-amp. Voltage dividers work for high-impedance inputs (ADCs, op-amp inputs) but can't deliver current. They're signal-level circuits, not power supplies.
Time Constants and Steady-State: What the Circuit Looks Like at t = 0 vs. t = ∞
Pure Ohm's law assumes steady state. V = IR is what you read after every capacitor has charged and every inductor has settled. At t = 0, the moment a switch flips, the picture is different. Capacitor voltage can't change instantly, so a discharged capacitor looks like a wire. Inductor current can't change instantly, so an empty inductor looks like an open circuit.
That gives you a quick rule for transient problems. At t = 0+, replace caps with shorts (if they were uncharged) and inductors with breaks (if they were de-energized). At t = ∞, replace caps with breaks (no DC current through a cap) and inductors with shorts (no DC voltage across an ideal inductor). The two limit cases bracket the answer, and the exponential tells you how the circuit gets from one to the other.
Even "pure resistor" circuits have a τ when you account for parasitics. Wire inductance (about 1 nH per millimeter of trace) and stray capacitance (a few pF between adjacent traces) form an unintended RLC network that rings on fast edges. For a 1 ns rise time, that's the regime where Ohm's law alone gives you wrong answers.
Steady-state shortcut: If a problem asks for the DC operating point and gives you a complicated network with caps and inductors, redraw with caps open and inductors shorted, then solve the resulting resistor network with Ohm's law and Kirchhoff. That's the answer at t = ∞.
Real Components and Tolerances (the ±10% Resistor Problem)
Resistors come in standard series. E12 (12 values per decade, ±10% tolerance) and E24 (24 values, ±5%) are the everyday options. Precision work uses E96 (±1%) or E192 (±0.1%). The ±10% tolerance bites harder than students expect: a nominal 1 kΩ resistor can read anywhere from 900 Ω to 1100 Ω, and the worst case in a divider stacks the tolerances.
| E12 Series (×1, ×10, ×100, ×1k, ×10k, ×100k, ×1M) | |||||
|---|---|---|---|---|---|
| 1.0 | 1.2 | 1.5 | 1.8 | 2.2 | 2.7 |
| 3.3 | 3.9 | 4.7 | 5.6 | 6.8 | 8.2 |
Tolerance isn't the only real-component issue. Temperature coefficient (TCR) shifts resistance with heat: a typical carbon-film part drifts ±200 ppm/°C, so a 10 kΩ resistor at 100 °C reads 10.16 kΩ. Metal-film hits ±50 ppm/°C and Vishay bulk-foil parts go to ±0.2 ppm/°C if you need it. Voltage coefficient and self-heating add a few more percent in extreme cases. For a current-limit resistor, none of this matters. For a precision divider feeding a 16-bit ADC, all of it does.
| Power Rating | Physical Size | Typical Use |
|---|---|---|
| 0.125 W | 0402 to 0603 SMD | Low-current signals |
| 0.25 W | 0805 SMD, 1/4 W axial | General purpose |
| 0.5 W | 1206 SMD, 1/2 W axial | LED circuits, drivers |
| 1 to 10 W | Power resistors, heatsink | Power supplies, loads |
Power rating is the other constraint. A 100 Ω resistor with 0.1 A through it dissipates P = (0.1)² × 100 = 1 W. A 0.25 W part would overheat and fail. Use a resistor rated at least 2× your calculated power. If P = 0.3 W, pick a 0.5 W or 1 W resistor for margin.
Rounding rule: When your calculated value isn't standard, round to the nearest. For LED current limiting, round up (higher R, lower current, safer LED). For voltage dividers, small deviations matter less. Pick the closest E-series value and check the resulting V_out against your tolerance budget.
AC vs. DC: Knowing Which Equations Apply
For DC, V = IR is the whole story. For AC you replace R with impedance Z, a complex number that captures both magnitude and phase. A pure resistor has Z = R (real, no phase shift). A capacitor has Z = 1/(jωC), which decreases with frequency. An inductor has Z = jωL, which increases with frequency. The j is just √(−1), the engineering name for the imaginary unit.
That means at 60 Hz mains, a 10 µF capacitor presents |Z| = 1/(2π × 60 × 10⁻⁵) ≈ 265 Ω of impedance, and the current through it leads the voltage by 90°. Treat that capacitor as a resistor and your numbers will be off in both magnitude and timing. Run the same cap at 1 MHz and |Z| drops to 16 mΩ, essentially a short.
For purely resistive AC loads (incandescent bulbs, heaters, kettles), V = IR works using RMS values: 120 V_rms across a 12 Ω element gives 10 A_rms and 1200 W. Reactive loads need power factor: P_real = V_rms × I_rms × cos φ. A motor with PF = 0.8 draws 1250 VA of apparent power for every 1000 W of real power. The wiring sizes for the apparent power, the meter charges for the real power.
Rule: For non-ohmic devices (LEDs, diodes, transistors, thermistors), use their characteristic curves or datasheet specs, not Ohm's law. Apply Ohm's law only to the resistors in the circuit. An LED below its forward voltage (about 2 V for red, 3.2 V for blue) draws zero current. Above that threshold, current shoots up exponentially with voltage. There's no single R to plug in. That's why LED circuits use a current-limiting resistor: the resistor obeys Ohm's law, and the LED is treated as a fixed forward voltage drop.
Worked Example: 12 V Car Battery Driving a 4.7 Ω Headlight
A typical halogen H4 headlight bulb has a hot resistance around 4.7 Ω at full operating temperature. Drive it from a fully charged 12.6 V lead-acid battery and walk through the numbers. Then add the battery's internal resistance and watch what changes.
Ideal battery (zero internal resistance)
Step 1: Current
I = V / R = 12.6 / 4.7 = 2.68 A
Step 2: Power dissipated by the bulb
P = V × I = 12.6 × 2.68 = 33.8 W
Cross-check with P = V²/R = 12.6² / 4.7 = 33.8 W. Same answer. Good.
Step 3: Energy per hour
E = P × t = 33.8 W × 3600 s = 121,680 J ≈ 122 kJ ≈ 0.0338 kWh
Running both headlights for an hour pulls about 67.6 Wh from the battery. A 60 Ah lead-acid at 12 V stores roughly 720 Wh, so the headlights alone draw it down by about 9% per hour.
Now with internal resistance R_int = 0.05 Ω
I = V_oc / (R_int + R_bulb) = 12.6 / (0.05 + 4.7) = 12.6 / 4.75 = 2.65 A
Terminal voltage drops to V_term = 12.6 − I × R_int = 12.6 − 2.65 × 0.05 = 12.47 V. The bulb sees 12.47 V instead of 12.6 V. Power dissipated in the bulb drops to 12.47 × 2.65 = 33.0 W. The 0.05 Ω of internal resistance converts about 0.35 W into heat inside the battery.
Lesson: a healthy lead-acid battery has internal resistance under 10 mΩ, so the headlight loss is negligible. A tired battery climbs to 100 mΩ or more, dropping terminal voltage and dimming the lights even though the open-circuit voltage still measures 12.6 V on a multimeter. That's why a load-test (cranking the engine while watching the voltage) catches a battery your no-load reading misses.
References
For derivations, design rules, and worked examples, the references below are what working engineers actually open. Horowitz & Hill is the bench reference. Sedra & Smith is the academic standard. The web references give you searchable derivations when you don't want to flip through a 1200-page book.
- Horowitz, P. & Hill, W. (2015). The Art of Electronics (3rd ed.). Cambridge University Press. Chapters 1 and 2 cover DC circuits, voltage dividers, and the practical limits of Ohm's law.
- Sedra, A. S. & Smith, K. C. (2020). Microelectronic Circuits (8th ed.). Oxford University Press. Linear-network analysis and the small-signal model live here.
- All About Circuits DC Circuit Theory. Free online textbook with worked examples.
- NIST SI unit definitions for volt, ampere, ohm, and watt: physics.nist.gov/cuu/Units.
- Vishay application note AN-200 on resistor temperature coefficient and stability for precision applications.
Limitations and Assumptions
- •Ohmic devices only: resistors, wires, heaters. Not LEDs, diodes, transistors, or thermistors above their nonlinear region.
- •DC or low-frequency AC into resistive loads: for general AC, replace R with impedance Z and account for phase.
- •Ideal components: no wire resistance, contact resistance, or temperature drift. Add tolerances explicitly when they matter.
- •Steady state: transients, capacitance, and inductance aren't modeled here. See the RC and RLC calculators for those.
Debugging V-I-R-P Calculation Problems
Real questions from hobbyists stuck on resistor values, voltage drops, and power ratings.
What is Ohm's law and how is it used?
Ohm's law states that the voltage across a resistor equals the current through it times its resistance: V = IR. Voltage in volts, current in amperes, resistance in ohms. Rearranged, you get I = V/R for current and R = V/I for resistance. Pick whichever form excludes the variable you don't have. A 12 V battery driving 100 mA through a circuit means R = 12/0.1 = 120 Ω. A 5 V supply across a 220 Ω resistor pushes I = 5/220 = 22.7 mA. The law also gives power three different ways: P = VI = I²R = V²/R. A 220 Ω resistor with 5 V across it dissipates 5²/220 = 0.114 W. That fits inside a 0.25 W resistor's rating with margin to spare. Ohm's law is exact for ohmic materials, where R doesn't change with V or I. Most metallic resistors and doped semiconductors at low fields are ohmic. Diodes, transistors, plasma columns, and gas discharges aren't, so the equation only describes part of the I-V curve. For non-ohmic devices, you need the full I-V characteristic. The most common errors aren't in the math. They come from real-world effects the equation ignores: internal battery resistance, wire voltage drops, loading by attached circuits, and resistor tolerance bands.
My 9V battery measured 9.2V with no load but drops to 7.8V when I connect my circuit—what's happening?
That's internal resistance at work. Every battery has resistance inside it (typically 1-5 Ω for a 9V battery, more as it ages). When current flows, some voltage drops across that internal resistance: V_drop = I × R_internal. If your circuit draws 300 mA and internal resistance is 4 Ω, you lose 1.2V inside the battery itself, leaving 8V at the terminals. This gets worse as batteries discharge—internal resistance climbs, so voltage sag under load increases. Always measure voltage while the circuit is running, not open-circuit, to get the actual working voltage. If your circuit needs stable 9V, use a voltage regulator.
I calculated 150 Ω for my LED but only have 220 Ω—will the LED still light up?
Yes, it will light—just dimmer. Using 220 Ω instead of 150 Ω reduces current (I = V/R), so less current flows through the LED. For a typical indicator LED rated at 20 mA, going from 150 Ω to 220 Ω might drop current from 20 mA to ~14 mA. The LED works fine; brightness drops maybe 25-30%. This is actually safer than running at max current—the LED lasts longer and runs cooler. The only time it's a real problem is if you go too high (say 1 kΩ) and the LED is barely visible, or if you're driving high-power LEDs that need minimum current for proper operation.
I'm getting 0.47 Ω from my multimeter but the resistor is marked 470 Ω—did I break it?
Nine times out of ten, this comes from measuring in-circuit. When a resistor is connected to other components, your multimeter measures the parallel combination of that resistor with everything else connected to it. If a 470 Ω resistor is in parallel with another path that has ~0.5 Ω equivalent (like a transistor base-emitter junction or another low-value resistor), you'll read the lower value. Always remove at least one leg of the resistor from the circuit before measuring. Also check your meter's range—if it's auto-ranging and jumped to a different scale, you might be reading 0.47 kΩ = 470 Ω, which would be correct.
The calculator says I need 33.3 Ω but that's not a standard value—what do I actually buy?
Grab 33 Ω from the E24 series (the most common). Standard resistor values follow the E-series: E12 has 12 values per decade (10, 12, 15, 18, 22, 27, 33, 39, 47, 56, 68, 82), E24 has 24 values, E96 has 96. For current-limiting and basic circuits, rounding to the nearest standard value works fine—5% difference is within typical resistor tolerance anyway. If you need exactly 33.3 Ω, you can series two resistors (33 Ω + 0.33 Ω ≈ not practical) or parallel two (two 68 Ω in parallel = 34 Ω). Usually, just use 33 Ω and your LED or load won't notice the 1% difference.
I paralleled two 100 Ω resistors expecting 50 Ω but I'm measuring 48 Ω—are they bad?
They're fine—that's normal tolerance. If both resistors are 5% tolerance, each could be anywhere from 95-105 Ω. Two 96 Ω resistors in parallel give 48 Ω, which is exactly what you're seeing. For precision work, use 1% tolerance resistors (brown band instead of gold). Or measure individual resistors before paralleling and do the math: 1/R_total = 1/R1 + 1/R2. Your 48 Ω result just means both resistors happened to be on the low side of their tolerance band. This is why production circuits use measured values for calibration, not nominal values.
Why does my voltage divider output 2.1V instead of the calculated 2.5V when I connect my sensor?
Loading effect. Voltage dividers assume nothing is connected to the output (high-impedance load). When your sensor draws current, it acts like a resistor in parallel with R2, changing the effective bottom resistor and dropping output voltage. If your divider uses 10 kΩ resistors and your sensor has 10 kΩ input impedance, the effective R2 becomes 5 kΩ (10k || 10k), completely changing the ratio. Fix: use resistor values at least 10× lower than your load impedance, or buffer the output with an op-amp voltage follower. For a sensor with 10 kΩ input, use 1 kΩ divider resistors so loading effect is only ~10%.
I'm using 50 feet of speaker wire for outdoor lights and they're dim—how do I calculate the problem?
Wire resistance is eating your voltage. 18 AWG wire is about 0.0065 Ω/ft, so 50 ft × 2 (round trip) = 100 ft × 0.0065 = 0.65 Ω. If your lights draw 5A total, that's V_drop = 5 × 0.65 = 3.25V lost in the wire. On a 12V system, you're delivering only 8.75V to the lights—no wonder they're dim. Solutions: use thicker wire (14 AWG = 0.0025 Ω/ft, cuts drop to 1.25V), increase source voltage to compensate, or run higher voltage and step down at the lights. For 12V DC runs, keep wire loss under 3% (0.36V)—that requires 10 AWG for 50 ft at 5A.
My resistor is getting hot—how do I know if it's about to fail?
Calculate power dissipation: P = I²R or P = V²/R. If your 1 kΩ resistor has 15V across it, P = 225/1000 = 0.225W. A standard 0.25W resistor runs warm at 0.225W (90% of rating) but survives. At 0.3W it overheats. Touch test: too hot to touch comfortably (>50°C) means it's stressed. Visible discoloration (brown burn marks) means damage is occurring. Rule of thumb: derate by 50%—use 0.5W resistors for 0.25W loads. They run cooler, last longer, and handle transients better. If your resistor is already browning, replace it with higher wattage immediately before it fails open and potentially damages other components.
Can I use Ohm's Law to calculate the resistor for my 3W LED or is that different?
Ohm's Law applies but power LEDs are current-controlled, not voltage-controlled. The calculation is the same: R = (V_supply - V_LED) / I_LED. For a 3W LED (typically 3.2V forward, 700-1000 mA), with 12V supply: R = (12 - 3.2) / 0.7 = 12.6 Ω. But here's the catch—power dissipation in that resistor: P = 8.8V × 0.7A = 6.2W. You'd need a big, expensive, hot resistor. That's why power LEDs use constant-current drivers instead—they're switching regulators that efficiently provide exactly 700 mA regardless of input voltage, wasting minimal power as heat. Use resistors for indicator LEDs (20 mA), drivers for power LEDs (350+ mA).
My multimeter shows different readings when I reverse the probes on a resistor—is that normal?
No, resistors are bidirectional—readings should be identical either direction. If you see different values, you're probably measuring something that's NOT a pure resistor: a diode junction in parallel (common on circuit boards), a capacitor that hasn't fully discharged, or corroded/dirty contacts affecting the measurement. Try this: remove the component from the circuit completely, clean the leads with isopropyl alcohol, and measure again. If readings still differ, it's not a resistor or there's a parallel diode you missed. True resistors made of carbon, metal film, or wire-wound are purely ohmic—current flows equally well in both directions with identical resistance.