Ever wondered why a tiny battery can light up a whole room of LEDs? The secret lies in a simple rule called Ohm's Law.
Ohm's Law tells us that the electric pressure (voltage) pushing charges through a wire is directly linked to how many charges flow (current) and how much the wire resists that flow (resistance). In short, V = I × R.
What is Ohm's Law?
Voltage (V) is like the water pressure in a hose – it tries to push the water (or electric charge) forward. Current (I) is the amount of water that actually moves through the hose each second, measured in amperes (A). Resistance (R) is anything that narrows the hose, making it harder for water to flow, measured in ohms (Ω).
Georg Ohm, a German physicist, discovered that if you keep the resistance fixed, the voltage and current change together in a straight‑line relationship. That line is the famous formula V = I × R.
How to Use the Formula
Whenever you know any two of the three quantities, you can find the third. Just rearrange the equation:
- To find current: I = V / R
- To find voltage: V = I × R
- To find resistance: R = V / I
Worked Example 1 – Finding Current
Suppose a 12 V battery is connected to a resistor of 4 Ω. What is the current?
Using I = V / R, we get I = 12 V / 4 Ω = 3 A. So 3 amperes of charge flow each second.
Worked Example 2 – Series Circuit
Two resistors, 2 Ω and 3 Ω, are placed one after the other (in series) across a 15 V supply. Find the total current.
In a series circuit, resistances add up: R_total = 2 Ω + 3 Ω = 5 Ω. Now I = V / R_total = 15 V / 5 Ω = 3 A. The same 3 A passes through both resistors.
Worked Example 3 – Parallel Circuit
Two resistors, 6 Ω and 3 Ω, are connected side‑by‑side (in parallel) to a 12 V battery. What is the current through each resistor?
In parallel, each resistor sees the full voltage. So:
- I₁ = 12 V / 6 Ω = 2 A
- I₂ = 12 V / 3 Ω = 4 A
The total current supplied by the battery is I_total = 2 A + 4 A = 6 A.
Quick Comparison Table
| Quantity | Symbol | Unit | What it means |
|---|---|---|---|
| Voltage | V | Volt (V) | Electric pressure pushing charges |
| Current | I | Ampere (A) | Flow of charge per second |
| Resistance | R | Ohm (Ω) | Opposition to charge flow |
Step‑by‑Step Guide to Solve Simple Circuit Problems
Common Mistakes to Avoid
- Mixing up the symbols – V is voltage, not velocity.
- Forgetting that resistances in series add, while in parallel they combine differently.
- Leaving the answer in kilo‑ohms or milliamps without converting to base units.
📝 Likely Exam Questions
- Question: A 9 V battery is connected to a resistor of 3 Ω. Find the current and the power dissipated (P = V × I).
Answer: I = 9 V / 3 Ω = 3 A. Power P = 9 V × 3 A = 27 W. - Question: Two resistors of 4 Ω and 6 Ω are in series across a 24 V supply. What is the voltage across each resistor?
Answer: Total R = 10 Ω, I = 24 V / 10 Ω = 2.4 A. V₁ = I × 4 Ω = 9.6 V, V₂ = I × 6 Ω = 14.4 V. - Question: In a parallel circuit, a 12 V source feeds a 2 Ω resistor and an unknown resistor. The total current drawn is 9 A. Find the unknown resistance.
Answer: I₁ = 12 V / 2 Ω = 6 A. Remaining current for unknown resistor: I₂ = 9 A – 6 A = 3 A. R₂ = 12 V / 3 A = 4 Ω. - Question: State Ohm's Law and explain its significance in everyday appliances.
Answer: Ohm's Law: V = I × R. It tells us how voltage, current, and resistance are linked, letting engineers design safe lighting, heaters, and chargers. - Question: A circuit has a 5 Ω resistor and a 10 Ω resistor in series with a 30 V battery. What is the total resistance and the current?
Answer: R_total = 5 Ω + 10 Ω = 15 Ω. I = 30 V / 15 Ω = 2 A.