Ohm's Law Simulator: V = IR Circuit with Examples | Shikkha Web

Conventional currentHeat in the resistorCurrent resistanceEarlier resistance

Speed

Controls

Battery voltage, V12.0 V

Resistance, R₁4 Ω

Circuit

Show electron flow

Readings

Potential difference, V: 12V
Equivalent resistance, R: 4.00Ω
Current, I: 3.00A
Power, P = VI: 36.0W

How to use this simulation

Move the voltage slider: with the resistance fixed, the ammeter reading rises and falls in the same proportion.
Raise the resistance: the charges slow down, the current drops, and the V–I line gets steeper. The slope is the resistance.
Choose "Series": both resistors carry the same current, but the voltage is shared between them.
Choose "Parallel": both branches get the full voltage, and charges race faster through the smaller resistance.

Before we start: Ohm's law is already in your home

Turn a ceiling fan's regulator and the fan slows down. Plug a phone into the wrong charger and it may refuse to charge, or get hot. When the power backup kicks in, the lights can look a little dimmer. Every one of these everyday moments comes down to one simple relationship: Ohm's law.

It sounds heavy, but it is really about three things: how hard we push the charges (voltage), how much charge actually flows (current), and how much the path fights back (resistance). Once you see how those three fit together, most circuit problems become quick arithmetic.

We will build it up from nothing: first the idea, then the law, then experiments in the simulation, then solved problems and exam practice. Take it one step at a time.

Understanding electricity with water pipes

A metal wire is packed with free electrons. Left alone they jiggle around randomly and go nowhere in particular. Connect a battery across the wire and they start drifting in one direction. That steady drift of charge is an electric current.

Now picture a water tank on a roof feeding a pipe. The higher the tank, the more pressure, and the harder the water pushes through. In a circuit, that pressure is the potential difference, or voltage. The battery is the pump that keeps the pressure up.

The amount of water passing a point each second is like current. In a circuit, current is the amount of charge (in coulombs) passing a point each second, measured in amperes.

If the pipe is narrow or clogged, less water gets through. That opposition is resistance. A thin, long or poorly conducting wire lets less current through for the same voltage.

So in plain words: more push means more flow, and more opposition means less flow. Ohm's law just says that in the language of maths.

The key words at a glance

Definitions and units come up in almost every exam, so it is worth learning this table properly.

Quantity	Symbol	Unit	Measured with	Plain meaning
Charge	Q	coulomb (C)	—	The stuff that flows
Current	I	ampere (A)	Ammeter (connected in series)	Charge per second, I = Q/t
Potential difference	V	volt (V)	Voltmeter (connected in parallel)	The "pressure" pushing charge
Resistance	R	ohm (Ω)	Ohmmeter	Opposition to the flow
Power	P	watt (W)	Wattmeter	Energy used per second

Ohm's law: the statement and the formula

Georg Simon Ohm, a German physicist, found by experiment that if a metal wire is kept at the same temperature, doubling the voltage across it doubles the current through it, tripling it triples the current, and so on.

The statement

The current through a conductor is directly proportional to the potential difference across its ends, provided the temperature and other physical conditions remain constant.

Never drop the "constant temperature" condition. When a wire heats up its resistance changes, and the simple proportion breaks down.

I ∝ Vat constant temperature

V = IRthe constant of proportionality R is the resistance

Three ways to write it

Know any two of V, I and R and you can find the third. For current, divide voltage by resistance; for resistance, divide voltage by current.

V = I × Rto find voltage

I = V ÷ Rto find current

R = V ÷ Ito find resistance

The ohm

A conductor has a resistance of one ohm if a potential difference of one volt across it drives a current of one ampere through it: 1 Ω = 1 V/A.

What resistance depends on

A longer wire has more resistance, because the charges have further to go. A thicker wire has less, because there is more room to flow. And the material matters: copper, aluminium and nichrome all differ. That material property is called resistivity, ρ.

Say a wire has a resistance of 5 Ω. Double its length and it becomes 10 Ω; keep the length but double its cross-sectional area and it becomes 2.5 Ω.

Temperature matters too: most metals resist more when hot. A glowing bulb filament has far more resistance than the same filament when cold.

R = ρL / AL = length, A = cross-sectional area, ρ = resistivity

The V–I graph: a story told by a straight line

Keep the resistance fixed, raise the voltage step by step, and plot the current each time. The points fall on a straight line through the origin, because zero voltage means zero current.

With V on the vertical axis and I on the horizontal, the slope of that line is the resistance: a bigger resistance gives a steeper line. Swap the axes (I vertical) and the slope becomes 1/R, so a bigger resistance gives a flatter line. Always check which axis is which.

Conductors that give a straight line are called ohmic, such as a copper wire at steady temperature. Those that do not, such as a glowing filament or a diode, are non-ohmic and give a curve.

Five experiments to try in the simulation

Reading about it is one thing; seeing it is better. Try these one at a time and watch the readings panel each time.

Choose "One resistor" at 4 Ω. Raise the voltage from 12 V to 24 V: the current goes from 3 A to 6 A. Double the push, double the flow.
Keep the voltage fixed and raise the resistance. The moving charges slow down and the ammeter reading drops: more opposition, less flow.
Turn the voltage down to zero. Every charge stops and the current is zero. No push, no flow.
Choose "Series": charges move at the same pace through both resistors (same current), but the voltage splits, with more across the bigger one.
Choose "Parallel": the current splits between two branches, with more through the smaller resistance. Then switch on electron flow to see that electrons actually move the opposite way to conventional current.

Series and parallel combinations

Resistors joined end to end, one after another, are in series: the charge has only one path. Resistors joined across the same two points are in parallel: the charge splits between paths.

In series the resistances simply add, so the total goes up. In parallel you are adding extra paths, so the total goes down, below even the smallest single resistor.

Rs = R₁ + R₂equivalent resistance in series

1/Rp = 1/R₁ + 1/R₂equivalent resistance in parallel

Rp = R₁R₂ / (R₁ + R₂)shortcut for exactly two resistors

Feature	Series	Parallel
Current	Same through every resistor	Splits; more through the smaller resistance
Voltage	Shared in proportion to resistance	Same across every branch
Equivalent resistance	More than the largest	Less than the smallest
If one bulb fails	Everything goes off	The others stay on
Typical use	Decorative light strings, fuses	Home wiring

Solved problems, step by step

These use the same values the simulation starts with, so you can check every answer in the readings panel. Always write down "given" and "to find" first; it cuts mistakes in half.

Problem 1: current through one resistor

Given V = 12 V and R = 4 Ω. Find I.

I = V ÷ R = 12 ÷ 4 = 3 A. The power is P = VI = 12 × 3 = 36 W.

Problem 2: resistance from measurements

A wire with 6 V across it carries 0.5 A. What is its resistance?

R = V ÷ I = 6 ÷ 0.5 = 12 Ω. This is exactly how resistance is measured in the lab: read the voltmeter and the ammeter, then divide.

Problem 3: series combination

4 Ω and 6 Ω are joined in series across a 12 V battery.

Step 1: Rs = 4 + 6 = 10 Ω. Step 2: I = 12 ÷ 10 = 1.2 A. Step 3: V₁ = 1.2 × 4 = 4.8 V and V₂ = 1.2 × 6 = 7.2 V.

Check: 4.8 + 7.2 = 12 V, the battery voltage. Notice the bigger resistor takes the bigger share.

Problem 4: parallel combination

The same two resistors, now in parallel across the same 12 V battery.

Step 1: Rp = (4 × 6) ÷ (4 + 6) = 2.4 Ω. Step 2: total current I = 12 ÷ 2.4 = 5 A. Step 3: branch currents I₁ = 12 ÷ 4 = 3 A and I₂ = 12 ÷ 6 = 2 A.

Check: 3 + 2 = 5 A. The same battery pushes far more current in parallel, because the equivalent resistance is much smaller.

Problem 5: power, and a household bulb

In the series circuit, each resistor's power is P = I²R: P₁ = 1.2² × 4 = 5.76 W and P₂ = 1.2² × 6 = 8.64 W. In series, the bigger resistor gets hotter.

A bulb is rated 220 V, 100 W. Running normally, it draws I = P ÷ V = 0.45 A and has a resistance R = V² ÷ P = 484 Ω. Measure it cold and you will find much less, because the filament is non-ohmic.

P = VI = I²R = V²/Rthree forms of power, using Ohm's law

Mistakes almost everyone makes

Examiners see these again and again. Knowing them in advance is the easiest way to avoid them.

Leaving "at constant temperature" out of the statement.
Working out 1/Rp in a parallel circuit and forgetting to flip it to get Rp.
Connecting the ammeter in parallel and the voltmeter in series; it is the other way round.
Putting mA or kΩ straight into the formula without converting to A and Ω.
Assuming every conductor obeys Ohm's law; diodes and bulb filaments do not.
Thinking electrons flow the same way as conventional current; conventional current goes from + to −, electrons the opposite way.

Exam corner: questions you will meet

These are the shapes the question usually takes, with the short answer an examiner wants.

State Ohm's law. At constant temperature, the current through a conductor is proportional to the potential difference across it; V = IR.
Define one ohm. The resistance of a conductor that carries one ampere when one volt is applied across it.
Draw and explain the V–I graph for an ohmic conductor. A straight line through the origin; its slope (V against I) is the resistance.
Why is the equivalent resistance in parallel smaller than any single resistor? Each extra branch adds a path for charge, so the total current rises for the same voltage.
Why are household appliances connected in parallel? Each gets the full mains voltage and can be operated independently.
Give an example of a non-ohmic device. A filament lamp or a diode; their V–I graphs are curved.

Quick revision

Read this list the night before an exam and the main ideas will come straight back.

V = IR, valid at constant temperature.
1 Ω = 1 V/A; resistance grows with length and falls with cross-sectional area, R = ρL/A.
The V–I graph is a straight line through the origin; with V against I, the slope is R.
Series: same current, Rs = R₁ + R₂. Parallel: same voltage, 1/Rp = 1/R₁ + 1/R₂.
P = VI = I²R = V²/R.
Ammeter in series, voltmeter in parallel.

Frequently asked questions

What is Ohm's law?

Ohm's law says the current through a conductor is directly proportional to the potential difference across it, as long as the temperature stays constant. It is written V = IR.

What is the unit of resistance?

The SI unit of resistance is the ohm (Ω). One ohm is the resistance that lets one ampere flow when one volt is applied: 1 Ω = 1 V/A.

What does a V–I graph show?

For an ohmic conductor it is a straight line through the origin. Plotting V against I, the slope equals the resistance; plotting I against V, the slope is 1/R.

Does Ohm's law apply to a diode or a light bulb?

No. A diode conducts far more easily one way than the other, and a bulb filament's resistance rises as it heats up, so neither gives a straight V–I line. They are non-ohmic.

How do you find the current if you know the voltage and resistance?

Divide the voltage by the resistance: I = V/R. For example, 12 V across 4 Ω gives 3 A.

Why is the total resistance lower in parallel?

Each branch is an extra path for the charge, like adding lanes to a road. More current flows for the same voltage, so the equivalent resistance is lower than the smallest resistor.

Why is an ammeter connected in series and a voltmeter in parallel?

An ammeter must measure all the current, so it goes in the current's path, in series. A voltmeter measures the difference between two points, so it connects across them, in parallel.