Physics · Current Electricity · NEET
All three are equal for a resistor, so any works if you know two quantities. The smart choice depends on what stays constant. In a SERIES circuit the current I is the same in every resistor, so use P = I²R (bigger R gets more power). In a PARALLEL circuit the voltage V is the same across each resistor, so use P = V²/R (smaller R gets more power). This one habit saves you in most NEET problems.
They are the same number for a given resistor at a given instant. They only look opposite because we ask 'what if we change the circuit?'. When current is fixed and you compare resistors, P = I²R says power rises with R. When voltage is fixed, P = V²/R says power falls with R. Same formula family, different quantity held constant, so use the form whose fixed quantity you actually know.
A heater has a fixed resistance R (set by its coil). Its rating uses P = V²/R. If V falls, R does not change, so P = V²/R falls as the square of voltage. For a 400 W, 220 V heater run at 200 V: P₂ = (200/220)² × 400 ≈ 331 W. This is a very common NEET numerical, so remember P ∝ V² at fixed R.
For bulbs rated at the SAME voltage, higher power means LOWER resistance, because R = V²/P. A 100 W, 220 V bulb has less resistance than a 60 W, 220 V bulb. This flips the intuition many students carry, so always convert a rating to resistance using R = V²/P before comparing.
Power P is the rate of using energy (unit watt, W = J/s). Energy is power multiplied by time: E = P × t = VIt = I²Rt = (V²/R)t (unit joule). Your electricity bill measures energy in kilowatt-hour: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. NEET may ask for either, so read whether the question wants watts or joules.
A filament bulb (500 W, 100 V) is to be used in a 230 V supply. A resistance R is connected in series so it works perfectly at 500 W. The value of R is:
A room heater is rated 400 W, 220 V. If the supply voltage drops to 200 V, the power consumed (approximately) is:
Two resistors 100 Ω and 200 Ω are connected in parallel. The ratio of thermal energy developed in 100 Ω to that in 200 Ω in a given time is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Power is in watt (W), where 1 W = 1 J/s = 1 volt-ampere. Energy is in joule (J), and commercially in kilowatt-hour (kWh), with 1 kWh = 3.6 × 10⁶ J.
P = VI is always valid for any device. P = I²R and P = V²/R apply to a purely resistive element (ohmic), where V = IR holds. For such a resistor all three give the same power.
Energy (kWh) = Power (kW) × time (hours). For example, a 2000 W heater for 3 hours uses 2 kW × 3 h = 6 kWh = 6 units on the bill.
Power lost in the wires is P_loss = I²R_wire. To send a fixed power P = VI at low current, engineers raise V so I is small, which cuts the I²R heat loss in long cables.
Yes. With V fixed, P = V²/R increases as R decreases, so a low-resistance appliance draws more current and more power.
