Science explained
Ohm's law and circuits explained simply, step by step
Ohm's law explained simply: V = IR, resistors in series and parallel, and how current and voltage behave in each, with fully worked circuit examples.
For students. By the Qlive team. . 5 min read.
This guide is for you if circuit questions feel like a jumble of formulas. In CBSE, Ohm’s law and resistors in series and parallel are taught in Class 10 in the Electricity chapter. They are also core topics in ICSE Class 10, GCSE and IGCSE physics. Learn three ideas well and most circuit questions follow from them.
The idea in plain words
Three quantities describe a simple circuit:
| Quantity | Symbol | Unit | What it means |
|---|---|---|---|
| Current | I | ampere (A) | How much charge flows past a point each second |
| Potential difference (voltage) | V | volt (V) | The energy given to or used by each unit of charge |
| Resistance | R | ohm (Ω) | How much a component opposes the current |
Ohm’s law says that, for a conductor at constant temperature, the current through it is directly proportional to the potential difference across it:
V = I × R
Rearranged: I = V/R and R = V/I.
A useful picture: voltage is the push, resistance is how hard it is to get through, and current is how much flows as a result.
Series and parallel circuits
| Series (one loop) | Parallel (separate branches) | |
|---|---|---|
| Current | Same everywhere | Splits between branches, branch currents add to the total |
| Voltage | Shared between components, adds to the supply voltage | Same across each branch |
| Total resistance | R = R₁ + R₂ + R₃ … | 1/R = 1/R₁ + 1/R₂ + 1/R₃ … |
| Effect of adding a resistor | Total resistance goes up | Total resistance goes down |
Worked examples
Example 1: using V = IR
A 4 Ω resistor is connected to a 12 V supply. Find the current.
- I = V/R
- I = 12 ÷ 4
- I = 3
Answer: 3 A
Example 2: resistors in series
A 2 Ω resistor and a 4 Ω resistor are connected in series to a 6 V battery. Find the current and the voltage across each resistor.
- Total resistance: R = 2 + 4 = 6 Ω
- Current: I = V/R = 6 ÷ 6 = 1 A. In series, this current flows through both resistors.
- Voltage across the 2 Ω resistor: V = IR = 1 × 2 = 2 V
- Voltage across the 4 Ω resistor: V = 1 × 4 = 4 V
Answer: current 1 A; 2 V across the 2 Ω resistor and 4 V across the 4 Ω resistor
Check: 2 + 4 = 6 V, the supply voltage.
Example 3: resistors in parallel
A 3 Ω resistor and a 6 Ω resistor are connected in parallel to a 6 V supply. Find the total resistance, the total current and the current in each branch.
- 1/R = 1/3 + 1/6
- Use a common denominator: 1/R = 2/6 + 1/6 = 3/6 = 1/2
- Flip it: R = 2 Ω
- Total current: I = V/R = 6 ÷ 2 = 3 A
- Each branch has the full 6 V across it:
- 3 Ω branch: I = 6 ÷ 3 = 2 A
- 6 Ω branch: I = 6 ÷ 6 = 1 A
Answer: total resistance 2 Ω, total current 3 A, branch currents 2 A and 1 A
Check: 2 + 1 = 3 A. Also, 2 Ω is smaller than both 3 Ω and 6 Ω, as it should be in parallel.
Example 4: a combined circuit
A 2 Ω resistor is connected in series with a parallel pair of 6 Ω and 3 Ω resistors. The supply is 12 V. Find the current from the supply and the current through the 6 Ω resistor.
- Parallel pair: 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R = 2 Ω
- Total resistance: 2 + 2 = 4 Ω
- Supply current: I = 12 ÷ 4 = 3 A
- Voltage across the series 2 Ω resistor: V = 3 × 2 = 6 V
- That leaves 12 - 6 = 6 V across the parallel pair
- Current through the 6 Ω resistor: I = 6 ÷ 6 = 1 A (and through the 3 Ω resistor: 6 ÷ 3 = 2 A)
Answer: supply current 3 A; current through the 6 Ω resistor is 1 A
Check: 1 + 2 = 3 A, matching the supply current.
Common mistakes and how to avoid them
Adding parallel resistors as if they were in series. 3 Ω and 6 Ω in parallel is 2 Ω, not 9 Ω. If your parallel total is bigger than the smallest resistor, something has gone wrong.
Forgetting to flip at the end. If 1/R = 1/2, then R = 2 Ω, not 0.5 Ω. Write “1/R =” clearly so you remember the last step.
Unit slips. Milliamps must be changed to amps before using V = IR. 250 mA = 0.25 A. Similarly, 2 kΩ = 2,000 Ω.
Using the supply voltage for one component in series. In series the voltage is shared. Use the voltage across that particular component.
Thinking current gets used up. In a series circuit the current is the same all the way round. Energy is transferred to components, but the charge keeps flowing.
Connecting meters the wrong way. Ammeter in series, voltmeter in parallel.
Try these
- A current of 2 A flows through a 5 Ω resistor. Find the potential difference across it.
- A lamp on a 230 V supply draws a current of 0.5 A. Find its resistance.
- Resistors of 3 Ω, 5 Ω and 4 Ω are in series with a 24 V supply. Find the current.
- Find the total resistance of a 4 Ω and a 12 Ω resistor in parallel.
- Two 10 Ω resistors are connected in parallel to a 10 V supply. Find the total current and the current in each resistor.
Answers
- V = IR = 2 × 5 = 10 V.
- R = V/I = 230 ÷ 0.5 = 460 Ω.
- R = 3 + 5 + 4 = 12 Ω, so I = 24 ÷ 12 = 2 A.
- 1/R = 1/4 + 1/12 = 3/12 + 1/12 = 4/12 = 1/3, so R = 3 Ω.
- 1/R = 1/10 + 1/10 = 2/10, so R = 5 Ω. Total current = 10 ÷ 5 = 2 A, and 1 A through each resistor.
Still unsure?
Circuit questions use fractions and rearranging formulas as much as physics, so the gap is not always where it seems. A Qlive teacher can work through it with you one-to-one, starting with a free assessment that finds the exact gap. You can read more about how we teach on our physics page and our CBSE page.