Ohm's Law and Electrical Power — V, I, R and P Connected
Electricity questions in Class 10 and Class 12 look varied — bulbs, heaters, fuses, electricity bills, series and parallel networks — but almost all of them come from just four letters: V, I, R and P. Once you can move freely between Ohm's law and the three power formulas, most numerical questions become one or two lines of work. This article shows the connections, then proves them on four typical exam problems.
Ohm's law
| Symbol | Quantity | SI unit |
|---|---|---|
| V | Potential difference across the component | volt (V) |
| I | Current through the component | ampere (A) |
| R | Resistance of the component | ohm (Ω) |
| P | Electrical power — energy converted per second | watt (W) |
Ohm's law says that for a metallic conductor at constant temperature, the current is directly proportional to the potential difference. That temperature condition is part of the law, not a footnote — a filament lamp gets hotter as it runs, its resistance rises, and its V–I graph bends. Such devices are called non-ohmic.
The power triangle — three formulas, one idea
Power is the rate at which electrical energy is converted into heat, light or motion. The starting definition is P = VI. Substituting Ohm's law into it gives two more useful forms:
P = I² R (put V = IR into P = VI)
P = V² / R (put I = V/R into P = VI)
All three give the same answer for the same component. You choose whichever one uses the two quantities you actually know — exactly the "missing variable" idea that works in kinematics. If you know current and resistance, use I²R. If you know voltage and resistance, use V²/R. Never convert unnecessarily.
Energy follows immediately, because energy = power × time:
E (kWh) = P (kW) × t (hours) — the "unit" on your electricity bill
One kilowatt-hour is 1000 W × 3600 s = 3.6 × 10⁶ J.
Worked example 1 — the basic substitution
Question: A 4 Ω resistor is connected across a 12 V battery. Find the current and the power dissipated.
Current: I = V / R = 12 / 4 = 3 A
Power, using P = VI: P = (12)(3) = 36 W
Check with the other two forms:
P = I²R = (3)²(4) = (9)(4) = 36 W ✔
P = V²/R = (12)² / 4 = 144 / 4 = 36 W ✔
All three agree, as they must. In an exam, doing one and checking with a second costs about ten seconds and catches most arithmetic slips.
Worked example 2 — a bulb rated "60 W, 240 V"
Question: A bulb is marked 60 W, 240 V. Find the current it draws at its rated voltage and the resistance of its filament when working.
Current — we know P and V, so rearrange P = VI:
I = P / V = 60 / 240 = 0.25 A
Resistance — we know P and V, so use P = V²/R rearranged:
R = V² / P = (240)² / 60 = 57 600 / 60 = 960 Ω
Check: R = V / I = 240 / 0.25 = 960 Ω ✔
Important: a rating plate always states values at the rated voltage. Run the same bulb on 120 V and it will not give 30 W — its filament is cooler, so its resistance is lower than 960 Ω, and the simple V²/R calculation no longer describes it accurately.
Worked example 3 — a heater and the electricity bill
Question: An electric heater rated 1500 W runs on a 220 V supply. Find (a) the current, (b) its resistance, (c) the energy used in 2 hours in kWh, and (d) the cost at ₹8 per unit.
(a) I = P / V = 1500 / 220 = 6.82 A (to 2 decimals)
(b) R = V² / P = (220)² / 1500 = 48 400 / 1500 = 32.27 Ω
Check: R = V / I = 220 / 6.82 = 32.26 Ω — the small difference is only rounding. ✔
(c) P = 1500 W = 1.5 kW, so E = 1.5 × 2 = 3 kWh (3 units).
(d) Cost = 3 × 8 = ₹24
Notice the practical consequence of (a): a 6.82 A load must not sit on a 5 A socket. That is exactly the reasoning behind fuse-rating questions.
Worked example 4 — series vs parallel, and who gets hotter
Question: A 3 Ω and a 6 Ω resistor are connected across an 18 V supply, first in series and then in parallel. Find the power in each resistor in both cases.
Series. The same current flows through both, so use P = I²R.
Rtotal = 3 + 6 = 9 Ω
I = V / R = 18 / 9 = 2 A
P3Ω = I²R = (2)²(3) = 12 W
P6Ω = I²R = (2)²(6) = 24 W
Total = 12 + 24 = 36 W, and a direct check gives P = VI = (18)(2) = 36 W ✔
Parallel. Both have the full 18 V across them, so use P = V²/R.
P3Ω = (18)² / 3 = 324 / 3 = 108 W
P6Ω = (18)² / 6 = 324 / 6 = 54 W
Total = 162 W. Check: Req = (3 × 6)/(3 + 6) = 18/9 = 2 Ω, so
P = V²/Req = 324 / 2 = 162 W ✔
The result worth memorising: in series the larger resistance dissipates more power; in parallel the smaller resistance dissipates more. This is a very common one-mark conceptual question, and picking the right power formula for the circuit type makes the answer obvious instead of something to memorise blindly.
Common mistakes that cost marks
- Using P = V²/R in a series circuit with the supply voltage. In series, each resistor gets only part of the supply voltage. Use P = I²R there, because the current is what is shared.
- Using P = I²R in a parallel circuit with the total current. Each branch carries a different current. Use P = V²/R there.
- Mixing kW and W with hours and seconds. kWh needs kilowatts and hours; joules need watts and seconds. 1500 W for 2 hours is 3 kWh, not 3000 J.
- Assuming a bulb's resistance is fixed. Filament resistance rises sharply with temperature, so a lamp is non-ohmic and its cold resistance is much lower than its working resistance.
- Forgetting that "unit" means kWh in Indian electricity bills. One unit is one kilowatt-hour.
- Squaring only one factor. In P = I²R only the current is squared; in P = V²/R only the voltage is squared. Writing (IR)² by mistake gives an answer that is R times too large.
Where this appears in exams
| Exam | Typical use |
|---|---|
| CBSE/ICSE Class 10 | Electricity chapter — heating effect, bulb ratings, electricity bills, fuse ratings |
| CBSE/ICSE Class 12 | Current electricity — internal resistance, Kirchhoff's rules, meter bridge |
| JEE Main & Advanced | Combination circuits, maximum power transfer, non-ohmic V–I graphs |
| NEET | Direct substitution questions on power and resistance combinations |
| GATE (engineering papers) | Network analysis built on the same V = IR foundation |
Verify every step instantly. Give the Ohm's Law tool any two of V, I, R and P and it returns the other two, using the correct power relation for the pair you entered — so you can confirm your I²R and V²/R answers match.
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